# *sigh* volatage divider....?

by WarPhalange
Tags: divider, sigh, volatage
 P: 343 I've been going over basic circuitry and got to the voltage divider. I don't understand the proof here: http://en.wikipedia.org/wiki/Voltage...r#General_case The first line makes sense. From Vi to Ground there are two resistors in series, so you add them, and V = IR and I is the same all around the whole circuit. Then the 2nd line I don't get. Why is it R2 and not R1? I thought current would go From Vi through R1 and then to Vo, not Ground to Vo.
 HW Helper P: 2,618 The potential difference across R2 (or Z2) is given by $V_{out} - 0$, where 0 is the ground voltage. For a resistor, V = IR so this means $V_{out} - 0 = IR_2[/tex] P: 343 Let me see if I have this right. Current goes from Vin through R1, which gives a voltage drop. Then since there is a branch, i.e. two things are in parallel, V across both is equal, but current probably isn't. So then you exploit the fact that at V at the branching point = Vout, and then you do this:  Quote by Defennder The potential difference across R2 (or Z2) is given by [itex]V_{out} - 0$, where 0 is the ground voltage. For a resistor, V = IR so this means [itex]V_{out} - 0 = IR_2[/tex]
I think I got it. Thanks a lot. :D

HW Helper
P: 2,618

## *sigh* volatage divider....?

Actually there isn't any "branch" in between R1 and R2. In fact I would say that if there is a branch there, the voltage divider principle does not work. It works only if the circuit elements are connected in series, which they would not be if there was a branch in between.
 P: 343 Sorry to revive this. I'm back in "electronics" mode and still stuck on this damn problem again... By branch I mean it branches off between R1 and R2 to Vout. I guess my next question is how do they know Iin = Iout? Wouldn't it be Vin = I_in(R1 + R2) and Vout = Iout*R2? Why do they use the same value for I? EDIT: Or are you assuming it's an open circuit, so that no current actually goes to the right into Vout? Then how would that work with a Load?
HW Helper
P: 2,618
The diagram is a little misleading. Vout does not denote a wire connected in between R1 and R2. It's just a line which indicates that the potential between the two resistors is V2. The same goes for Vin.

Iin = Iout is the assumption they make in order to derive the voltage divider principle. As said above, if that assumption is not true, ie. if there is a wire branching off from in between 2 resistors, then the potential divider principle does not work.

 Quote by WarPhalange EDIT: Or are you assuming it's an open circuit, so that no current actually goes to the right into Vout? Then how would that work with a Load?
EDIT: Yes you're right here. I don't know what you mean by "work with a load". It depends on whether you designate R2 as the load or something else.
 P: 343 Okay okay, that much makes sense. So how do you apply a voltage divider if you're not allowed to have anything between R1 and R2? I thought that was the whole point, where you kind of siphon off some of the current.
 HW Helper P: 2,618 The voltage divider principle is just a principle of circuits for easy evaluation of voltages across a circuit element, not a technique for getting a desired voltage out of something. To get a desired voltage across some complicated linear network, you first reduce it to it's Thevenin equivalent, then add resistors in the appropriate manner until you get the desired voltage.
 P: 343 I see... but if I had Rload at Vout >> R1 and R2, wouldn't that be approximately correct still? So if I used a voltmeter I could still read Vout from in between R1 and R2, right? I guess that's what confused me above all. I had thought this was something practical, not just a thought experiment type of thing.

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