The voltage is the force / electrical pressure, whereas the current (a quantity of n electrons) -- is the food for the circuit.

 

The 0.2 ohms internal resistance of the battery is in series with the 1 ohm load so the current is (10V/1.2 ohms=) 8.33A.
The 1 ohm load has a current of 8.33A so the voltage across it is (8.33A x 1 ohm=) 8.33V.

Nothing in the circuit has 50A. If you short the battery then 50A will try to heat it with (10V x 50A=) 500W!

 

Yeah that's right.

I recommend that you go and read some books or do a formal man.

 

Nothing in the circuit has 50A. If you short the battery then 50A will try to heat it with (10V x 50A=) 500W!

Forgetting maths and theory for a moment -- would it really be that much in practice?

I have never tried doing anything quite as stupid as that. I am getting dumber as I age though, so it is only a matter of time.

 

The .2Ω is in series with the 1Ω load.

Ohms laws states that 1 volt of electrical pressure will force 1 amp of current through a 1Ω load.

Hmm... So I assume that if I were to add another 1ohm resistor in series (R2), the amps the 10v would be be able to deliver to that second load (R2) would decrease even further...

 

If only I thought that you were genuinely looking for help I probably would answer.

 

The 0.2 ohms internal resistance of the battery is in series with the 1 ohm load so the current is (10V/1.2 ohms=) 8.33A.
The 1 ohm load has a current of 8.33A so the voltage across it is (8.33A x 1 ohm=) 8.33V.

So, from what I understand: figuring out the voltage at a load is on a load-by-load basis. So if I have two resistors, R1 and R2, each has a different voltage. Right?

But does each load have its own amperage? or is the amperage the same throughout the circuit?

For example, lets say I add another 1ohm resistor to that circuit in the link I mentioned. So now I have 2.2ohms series circuit resistance. Is the amperage 4.54A for both R1 and R2 or is R2 < R1 for Amps?

I would assume that R2 and R1 are equal. After all, if what you are saying is right (and I am sure it is) then 8.33A for 1ohm = 8.33V and the same would be true for another resistor whose existence in series lowers the amps of the circuit that the battery can provide.

 

If only I thought that you were genuinely looking for help I probably would answer.

If this was directed at me as per my comment above your post, then I was merely thinking out loud in that comment. But please correct me if you see an error in my thinking/logic

 

So, from what I understand: figuring out the voltage at a load is on a load-by-load basis. So if I have two resistors, R1 and R2, each has a different voltage. Right?

maybe. if they are the exact same value, then they should drop the same voltage. if for example, R1 is 2Ω, and R1 is 1Ω, then no, they will have different drops.

But does each load have its own amperage? or is the amperage the same throughout the circuit?

in a pure series circuit, the same amperage flows through all components. you calc this amperage by adding all the series resistances and then divide the voltage by that. for example if your R1 is 1Ω, R2 is 2Ω, and battery internal resistance is .2 ohms, then your total series resistance is 3.2Ω. if Rtot = 3.2Ω & voltage = 10V, then the amperage in all the circuit components is 3.125A.

so, if circuit current is 3.125A & battery internal resistance is .2Ω, then the voltage drop across the battery will be .625V (meaning only 9.375V read at the terminals, not as good as 10V, but not as bad as 8.3V)
R1:
E=I*R
E=3.125A*1Ω = 3.125V
R2:
E=I*R
E=3.125A*2Ω = 6.250V

notice that the sum of the voltage drops equals the battery voltage:
Ebatt + Er1 + Er2 = Etot
.625V + 3.125V + 6.250V = 10V

if we changed the value of R2 to 1Ω, then circuit amperage would be higher, and we would read less voltage at the battery terminals

If this was directed at me as per my comment above your post, then I was merely thinking out loud in that comment. But please correct me if you see an error in my thinking/logic

Please don't mind him. He's not a good representation of the community

 

I'm doing a project right now with a high drain lithium polymer flat cell that has 10mΩ DCR. With a full charge, it will dump over 400 amps on a hard short at the terminals. It's a fairly small battery too, about the size of a credit card. Those high drain LiPo cells are pretty amazing in how much power they can deliver.

 

400A into the 10mΩ internal resistance of the battery is 1600W!
A battery the size of a credit card will catch on fire very quickly with that much heating and how will you extinguish it? You don't.

 

400A into the 10mΩ internal resistance of the battery is 1600W!
A battery the size of a credit card will catch on fire very quickly with that much heating and how will you extinguish it? You don't.

true. I've not looked into the credit card sized LiPo, but I have been looking at what people are doing with the hobby "brick LiPo" (the kind used in model aircraft) and they routinely catch them on fire when precautions are not followed. Particular attention needs to be paid to the charge & discharge rates, the charge level, and the health of each individual cell. A bad cell can heat up and catch the battery on fire, and LiPo makes a relatively violent fire. Caution should be used. always charge outdoors, and it is recommended to charge inside a clay or ceramic planter pot, with a steel plate on top.

 

400A into the 10mΩ internal resistance of the battery is 1600W! A battery the size of a credit card will catch on fire very quickly with that much heating and how will you extinguish it? You don't.

Oh yea, it will pretty much explode into flames if you short it with a screwdriver or something. The cell has a 20C drain rating or 44A continuous. That would be the limit on what you can pull out of it. Still it can deliver max drain with a drop of less than a half volt.

It's the type of cell typically used in RC hobby batteries, but it's not the same config, it's an individual cell with solder tabs. The RC batteries are typically a stack of 2 or more series cells shrinkwrapped with the required connectors.

High drain LiPos are quite volatile in the event of a fault like a short or over-charge condition so you absolutely have to incorporate good reliable protection circuitry if you want to use one in a project.

The RC guys run these cells on the bleeding edge of charge and discharge rate so it's not uncommon for them to fail catasrophically. They also don't protect them. When protected and when running with a moderate charge/discharge rate, they're safe enough to use in a handheld project. They're actually pretty robust and it takes an extreme condition for them to go into thermal runaway.

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