Question about battery volatage

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bassmasta
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Re: Question about battery volatage

Post by bassmasta » Sat Feb 12, 2011 2:54 pm

DiscoPanda wrote:Nope. Let's simplify, say each load is a resistor. Say you have two identical resistors in series. Each is the same. Each will drop the same voltage, just like parallel identical resistors drop the same amperage. If you have a 10V source, with a 50 ohm, a 30 ohm, and a 20 ohm resistor in series, then the 20 ohm will drop 2V, the 30 ohm will drop 3V, and the 50 ohm will drop 5V. (((single resistor ohms / total series ohms) * voltage) = voltage dropped by single resistor).
I know what you're saying, but I'm just confused about the voltage at different points. Is this right?

Image
(positive is the top side of the battery, whoops)
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Re: Question about battery volatage

Post by DiscoPanda » Sat Feb 12, 2011 4:45 pm

bassmasta wrote:
DiscoPanda wrote:Nope. Let's simplify, say each load is a resistor. Say you have two identical resistors in series. Each is the same. Each will drop the same voltage, just like parallel identical resistors drop the same amperage. If you have a 10V source, with a 50 ohm, a 30 ohm, and a 20 ohm resistor in series, then the 20 ohm will drop 2V, the 30 ohm will drop 3V, and the 50 ohm will drop 5V. (((single resistor ohms / total series ohms) * voltage) = voltage dropped by single resistor).
I know what you're saying, but I'm just confused about the voltage at different points. Is this right?

http://i.imgur.com/Bsr5I.jpg
(positive is the top side of the battery, whoops)
Assuming that you mean connecting your meter between those points and ground gives those values, yes. Connecting the meter across A and B gives 4.5V, connecting the meter between B and C gives 4.5V, and connecting the meter between A and C gives 9V.

Now, the load you plan to connect to point B will need to have a negative/ground connection, right? Meaning it will be connected in parallel with R2. Remember the part about calculating parallel resistances? (1/((1/R2)+(1/Rload))) (which let's call R3 for convenience) is now the resistance in series with R1, not just R2 alone. Meaning that Vload=9V*(R3/(R1+R3))

And that's assuming that your load between B and negative/ground is a pure resistance load, no inductive load involved. Which probably isn't the case.

Like I said, there's a reason you don't see voltage dividers used more often - after all, a few resistors is cheaper than a voltage regulator, so if it was that easy everybody would be doing it that way.

(I can see how a voltage divider can be deceptively simple-looking though, I can remember when I was first learning all this stuff too... :D )

(Edit: Before someone points out that jumpering across a resistor and calculating the resistance of the pair would make you divide by zero - no. Wire has a resistance, it's simply very very tiny. Small enough that it's usually not important and is rounded to zero. There is such a thing as a wire-wound resistor made of a very long length of wire wound up into a cylinder. Generally smaller values, but capable of extremely tight tolerances on those values. And generally irrelevant to this discussion.)

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Re: Question about battery volatage

Post by face » Sun Feb 13, 2011 2:58 pm

read bens book, that should help
also this is what i learned in basic school physics so it cant be rocket science, just use your brains
in europe
u=I*R
I=U/R
R=U/I
parallel connection
(1/R)=(1/R1)+(1/Rn)
I=I1+In
U=U1=Un
the other connection which name i dont know in english
R=R1+Rn
I=I1=In
U=U1+Un
that way you can calculate the volts, amps etc
but if you want to regulate from 9 to 4,5 then use a resistor or devide them like the other connection
Image

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