Tonight

[happysteve]

http://youtu.be/Oz8RjPAD2Jk

 

[PEG]

Cheers,Tony,for a nice diversion from normality...been assigning currents with me sons coloured felt-tips,just need to pick a colour for voltages...:willy_nilly:

 

[Knobhead]

I'm going to use green.

I'll add another bit tonight.

 

[Knobhead]

I should have said at the start, there’s no correct method for this.

Now we have the total circuit current it’s possible to calculate the voltage drop across various parts. From here on everything is Ohms law.

The easiest point is the single 2Ω resistor.

Simple subtraction gives the volt drop for the network to the right.

Calculating the current through the 5Ω also give the current for the five resistor network, again by subtraction from the total current. This gives our 2[SUP]nd[/SUP] answer for point “E”

If we go back to this stage the double and triple networks are already calculated.

We already know the current through the five resistor network. The volt drop across the two network is calculated.

We now have all the volt drops across the network.

Once again Ohms law to the rescue.

Clear away all the rubbish that’s not needed.

As I said you may have another method to do the task.

 

[Knobhead]

Happysteve by any chance?

happysteve was the first with the answer.

 

[1stfix2ndfix]

Hi guys had a busy week/weekend but I'm going to have a look at all this help I'm being given in a min as it seems you've all been busy trying to help me which is very nice of you indeed.

 

[1stfix2ndfix]

Right well im still looking it. There was no way i would have know that.

 

[Knobhead]

Why? I’ve shown you how you break series/parallel networks in to small manageable chunks. Not everything can be solved in a single equasion.

 

[1stfix2ndfix]

Without the current at “A” nothing else can be calculated.

For parallel networks only two equations are needed R=1/((1/R[SUB]1[/SUB])+(1/R[SUB]2[/SUB])) for two resistors or R=1/((1/R[SUB]1[/SUB])+(1/R[SUB]2[/SUB])+(1/R[SUB]3[/SUB])) for three. Anything else the various networks are combined.

First calculate the two sub networks:

Add the two sub networks to give a single value:

We now have two parallel networks that have a resultant value:

Again by addition a total circuit resistance is given:

Good old Ohm’s law does the rest:

Its basic schoolboy maths and physics.






Or am I getting to old for this game?

the first one for 2.85


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


=0.35 ?


what am i doing wrong lol

 

[Simonslimline]

1/10 + 1/20 + 1/5 = 0.35

1/0.35 = 2.85.

 

[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