Tonight

[1stfix2ndfix]

Whos a silly boy.

now i can continue

 

[Knobhead]

the first one for 2.85


R= 1
-----------
1/10 + 1/20 + 1/5


=0.35 ?


what am i doing wrong lol

Not applying the inverse.

=1/((1/20)+(1/10)+(1/5))

Or as it appears on my screen

 

[Knobhead]

I've just been looking at series and parellel 1stfix, although on a 12V scale. I haven't come across a situation yet in the domestic setting though.

My way of understanding it on a basic level which is all I can take at the moment is :

View attachment 26416

They are all 1.5V batteries but depending if they are linked in parallel or series the voltage either increases or stays the same.

I may be wrong as I'm only just starting to look into this but :

1. In the above picture where the batteries are linked in parallel the total voltage is still 1.5V BUT the amount of Ah (amp hours) is increased. So if one battery has 3Ah (ie it can supply 3A for 1 hour), then if it is linked with the other batteries (4 in total - in parallel) the Ah is increased to 3Ah x 4 = 12Ah

2. If the batteries are linked in series you add the voltages together (so you get 6V) but the Ah stay the same at 3Ah in total.

If this is wrong I'm sure someone will say and I hope they do if it is as I would not want you getting misinformation.

I just read this and you’re wrong. Ahr doesn’t even enter in to series/parallel calculations.

4x3Ahr = 12Ahr no matter how you connect the batteries. Think of it as a gallon of water. You can pour it in to numerous containers, it’s still a gallon of water.

The Ahr is the batteries capacity to store power.

 

[happysteve]

I just read this and you’re wrong. Ahr doesn’t even enter in to series/parallel calculations.

4x3Ahr = 12Ahr no matter how you connect the batteries. Think of it as a gallon of water. You can pour it in to numerous containers, it’s still a gallon of water.

The Ahr is the batteries capacity to store power.

Tony, I'm going to (respectfully) disagree. Steve (the other happy one) is correct. If you have 4x 3Ah 1.5V batteries connected in parallel, you have a 12Ah 1.5V battery. If you connect 4x 3Ah 1.5V batteries in series, you have a 3Ah 6V battery.

"Ah" (amp-hours) is sort of the battery's capacity to store power. Actually, we should be talking about energy. Batteries hold a fixed amount of energy... once that energy is gone, they're dead, and you either need to chuck 'em or recharge 'em.

So we should really be talking about the amount of energy stored in a battery, and we all know energy is measured in Joules (J).

We also know that Energy (in J) = Power (in W) x time (in s).

So 1 W of power consumed (or generated) for 1 second is 1 Joule of energy.

You could also called the Joule a "Watt second". A "Watt hour" (Wh) would be 1 x 60 x 60 J (3600J) and a "kilowatt hour" is 1000 x 60 x 60 = 3,600,000J = 3.6MJ.

Back to the batteries....

They are described as "3Ah" and we know they are 1.5V. So the total energy stored in each is 1.5 x 3 Wh = 4.5Wh (= 4.5 x 3600J = 16.2kJ). However you connect them up, the total amount of energy contained in 4 batteries is 4 x 4.5Wh = 18Wh. (4 batteries x 3Ah x 1.5V).

4 batteries in parallel, you have 1.5V. 18Wh / 1.5V = 12Ah.
4 batteries in series, you have 6V. 18Wh / 6 = 3Ah.

Hope this is helpful

 

[ImpededLoop]

If you have 4x 3Ah 1.5V batteries connected in parallel, you have a 12Ah 1.5V battery. If you connect 4x 3Ah 1.5V batteries in series, you have a 3Ah 6V battery.

Yes, I'd agree. Good explanation too

 

[Silly Sausage]

Back to the batteries....

They are described as "3Ah" and we know they are 1.5V. So the total energy stored in each is 1.5 x 3 Wh = 4.5Wh (= 4.5 x 3600J = 16.2kJ). However you connect them up, the total amount of energy contained in 4 batteries is 4 x 4.5Wh = 18Wh. (4 batteries x 3Ah x 1.5V).

4 batteries in parallel, you have 1.5V. 18Wh / 1.5V = 12Ah.
4 batteries in series, you have 6V. 18Wh / 6 = 3Ah.

Hope this is helpful

Yeah, but you're driving 4 times the current through a load with the batteries in series than parallel.
i.e. 4 x Power

The Ahr is the same for both configuration, just that the batteries will last 4 times longer when in parallel.

 

[happysteve]

Yeah, but you're driving 4 times the current through a load with the batteries in series than parallel.
i.e. 4 x Power

Oh aye, agreed (assuming the same resistive load).

The Ahr is the same for both configuration, just that the batteries will last 4 times longer when in parallel.

No, for the parallel configuration you've got 12Ah at 1.5V (=18Wh) and for the series configuration you've got 3Ah at 6V (=18Wh).

This confusion is just because the term "amp hour" isn't complete, as it assumes a certain voltage.

(Similar confusion arises when you say, "This is a 1kW [resistive] heater." It's only 1kW at the intended voltage (eg 230V). If you run it at 115V, it's only a 0.5kW heater.)

 

[davesparks]

And then of course the Ah value of a battery is only ever achieved at the specific test conditions that the battery was tested at!

If a battery is delivers 1A for 10 hours under test conditions then it is labelled as a 10Ah battery. But if you tried to get 10A out of that battery for an hour you would be sadly disappointed!

 

[happysteve]

Agreed, definitely. When I lived on a boat and relied on a bank of 12V batteries (3x110Ah 12V in parallel, plus a starter) I reckoned on roughly one third of that energy actually being useful (before the voltage dropped too much).

 

[Silly Sausage]

Oh aye, agreed (assuming the same resistive load).



No, for the parallel configuration you've got 12Ah at 1.5V (=18Wh) and for the series configuration you've got 3Ah at 6V (=18Wh).

This confusion is just because the term "amp hour" isn't complete, as it assumes a certain voltage.

(Similar confusion arises when you say, "This is a 1kW [resistive] heater." It's only 1kW at the intended voltage (eg 230V). If you run it at 115V, it's only a 0.5kW heater.)

'n' batteries have the same 'AmpHour' capacity whether they are connected in series or parallel.

I'll bung my proof up later!

It's a brave man that challenges Tony!!!

 

[happysteve]

'n' batteries have the same 'AmpHour' capacity whether they are connected in series or parallel.

Well of course each "cell" remains a 3Ah 1.5V battery. Unconnected you have four 3Ah 1.5V batteries. Connected in parallel, you have a 12Ah 1.5V battery. Connected in series, you have a 3Ah 6V battery.

I'll bung my proof up later!

I'm sure I'll agree with it. I think in your case (and Tony's) it's more about semantics than fundamental understanding.

It's a brave man that challenges Tony!!!

True dat.

Had to grow a pair some day, though...

 

[Silly Sausage]

Well of course each "cell" remains a 3Ah 1.5V battery. Unconnected you have four 3Ah 1.5V batteries. Connected in parallel, you have a 12Ah 1.5V battery. Connected in series, you have a 3Ah 6V battery.


I'm sure I'll agree with it. I think in your case (and Tony's) it's more about semantics than fundamental understanding.

True dat.

Had to grow a pair some day, though...

No, you'll have 12Ah 6V battery, it'll just drain in a 1/16th of the time of that of a 12Ah 1.5V battery (Power proportional to V^2), sameb load.
I was wrong in my 1st post saying 4 x the Power!

I think you're not taking account of time, possibly

It's good to stand up against his Lordship once in a while, keep him in check...slightly!!! :smilielol5:

 

[HandySparks]

(Similar confusion arises when you say, "This is a 1kW [resistive] heater." It's only 1kW at the intended voltage (eg 230V). If you run it at 115V, it's only a 0.5kW heater.)

Hi happysteve. I agree entirely with your reply to Tony (I thought about it and then decided to let it lie).

You might, however, care to correct your passing statement above.

 

[davesparks]

What's wrong with his statement about the 1kw heater?

 

[HandySparks]

What's wrong with his statement about the 1kw heater?

If you halve the voltage on a resistive load, the power reduces to a quarter. 1kW at 230V > 250W at 115V.

 

[davesparks]

Theoretically yes, but only if the temperature and resistance remains constant, however the temperature and therefore resistance will change and the change in power won't be quite so simple to calculate

 

[happysteve]

Brain mong!

 

[happysteve]

Yes, Handysparks (and also Davesparks), I have had a total brain mong here.

P = IV. P = V[SUP]2[/SUP]/R P = I[SUP]2[/SUP]R

If you halve the voltage, the current will halve (for fixed resistance). But the power will go down to a quarter of what it was.

Gosh, you have no idea how silly I feel right now. :redface:

 

[davesparks]

I have a very good idea how silly you feel, we all make these silly mistakes sometimes.

 

[HandySparks]

What's wrong with his statement about the 1kw heater?

Ask happysteve.

Brain mong!

No probs, it's a easy one to miss. :biggrin:

 

[Knobhead]

Before you all carry on bickering, have a look at two definitions AHr and AHrV.

 

[happysteve]

Before you all carry on bickering, have a look at two definitions AHr and AHrV.

From Wikipedia, the free encyclopedia:

-----
An ampere-hour or amp-hour (SI symbol A·h or A h; also denoted Ah) is a unit of electric charge, equal to the charge transferred by a steady current of one ampere flowing for one hour, or 3600 coulombs.
-----

Couldn't find "AhV" or "AhrV" but VAh is (assuming no reactance) Wh (Watt hour) which is a unit of energy, equal to 3.6kJ (see post #44).

I have an inkling of where you're going with this, but I'll let it play out...

So, let's remind the class of the problem:

We have four 1.5V batteries. Each battery has a sticker saying "3Ah" on the side.

I assert that if you connect these together in parallel, you have a 12Ah 1.5V battery.
The total energy stored in that battery is 12Ah x 1.5V = 18Wh = 18VAh (or 18AHrV if you like) = 64.8kJ.
The total electrical charge in that battery is (12 x 3600)C = 43,200C (coulombs).

As far as I'm aware, there isn't any disagreement there (please pipe up, anyone, if you disagree).

I also assert that if you connect these together in series, you have a 3Ah 6V battery.
The total energy stored in that battery is 3Ah x 6V = 18Wh = 18VAh (or 18AHrV if you like) = 64.8kJ.
The total electrical charge in that battery is (3 x 3600)C = 10,800C (coulombs).

Anyone disagree?

 

[HandySparks]

Before you all carry on bickering, have a look at two definitions AHr and AHrV.

No bickering Tony, just a few chaps keen to ensure that the basic physics here is correct, even while the regs, best practice, etc. are open to interpretation and debate.

Assuming that you're referring to "Ah" and "VAh":

Ah is a measure of electric charge. 1Ah is equal to 3600C (coulombs). 1C = 1A x 1s.

VAh is a measure of stored energy. More frequently referred to as Wh. 1Wh is equal to 3600J (joules). 1J = 1W x 1s.

Neither are a measure of power.


The above ignores any issues of change of capacity with discharge rate, change of voltage with charge, etc.

Happy to be corrected. Others may wish to elaborate on this.



Edit:
I see that happysteve has posted while I was mulling over my reply!