# Voltage divider

Gold Member
Hello,

I am confused about voltage division. In reference to the circuit on wikipedia, I don't understand why Vout = R2/(R1+R2)Vin and not R1/(R1+R2)Vin. The current goes through R1 first and has a drop over that resistor, not R2.

https://en.wikipedia.org/wiki/Voltage_divider#Resistive_divider

If you connect a lead BELOW R2, That would be your voltage divider measurement. I think that Wikipedia thought that the implication was an automatic assumption that readers would make.

Again the voltage measured is across R2. The diagram is just not showing the lead BELOW R2.

Look at this diagram. You could place the voltage over R1 as well, and then it would be R1/R1+R2.

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Something like this would be R1/R1+R2

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Gold Member
Why would I automatically assume where the lead is?

Wikipedia is a platform built of public contributions of information. Everyday people are helping to contribute information. Some people who have been working there butts off trying to study for big exams more than likely would have made that assumption. They would have known how to use a voltmeter to measure voltage across a resistor.

Beginners probably would not make that assumption, because they haven't toiled with it enough yet.

The author of that page probably SHOULD put the extra lead on the diagram. They make mistakes on Wiki sometimes. If you look on Wiki, you will find that you actually can contribute information.

Look in a big fat college Electronics textbook. You will see lots of them there.

berkeman
Mentor
Why would I automatically assume where the lead is?

Because the output voltage is with respect to ground.

So if the output voltage is with respect to the ground, then this image should be R1/R1+R2.

To the Mentor:

Is this correct for the voltage across R1?

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berkeman
Mentor
So if the output voltage is with respect to the ground, then this image should be R1/R1+R2.

To the Mentor:

Is this correct for the voltage across R1?

No. I haven't read the whole thread, so I apologize if I'm missing something. The traditional simple voltage divider circuit has R1 on top and R2 on the bottom. Vin it put into R1 with respect to ground, and Vout is across R2 with respect to ground. That's why the voltage divider equation is this:

$$V_o = V_i \frac{R_2}{R_1 + R_2}$$

FOIWATER
Gold Member
Maybe it could help if we show where the relationship comes from, then the OP can apply it to any configuration he likes, if he understand it - here it is attached

You asked how you know where the leads are.. you know which way current flows in the circuit. This derivation is kept with the polarities (for voltage drops) in mind which correspond to that. So, the voltage across R2 is measured with respect to the ground of the power supply. Does this make sense?

I genuinely hope this clarifies it for you

#### Attachments

• voltage divider.png
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What would the diagram look like if you used a voltmeter to measure the voltage across R1?

Traditionally, I know that it is measured across R2. I traditionally know the equation. So even I did what the author on Wikipedia did. I assumed you knew that you were suppose to multiply the ratio by the initial voltage to get the output.

But where would you place the voltmeter if you want to measure the voltage across R1?

Gold Member
I'm not following the equations, this still isn't so clear to me.

FOIWATER
Gold Member
OK please allow me to try again.

If I explain something which you already know, forgive me I am not trying to insult your intelligence.

It is implied in this circuit that the OTHER end of the voltage source is connected to ground. It is not shown here. So you could replace that ground symbol with a wire which forms a loop all the way around. The current flows out of the voltage source and around the circuit back to the source (in this case, to the ground since the source is implied grounded).

The current flowing through the resistor causes a voltage drop across the resistor. What does this mean in terms of the diagram? the voltage 'above' R1 is more positive than the voltage 'below' R1. the same is true for R2. In the equations I posted, the Voltage V2 is what a volt meter would read with the '+' lead 'above' R2 and the '-' lead 'below' R2 (grounded). The voltage divider rule is derived from that.

I am not sure, I may be wrong, but I believe some of the confusion stems from the fact of not knowing voltage polarities and which things are connected to what in the circuit.

Is there anything I can be more clear about? I would like to be able to help

Maylis,

Look at these diagrams. The wikipedia diagram DOESN'T show the bottom lead. The images that I posted have no IMPLICATIONS with respect to the traditional voltage divider equation. It shows the bottom lead. In most courses, you are suppose to see the bottom lead.

Vout = Vin(R2/R1+R2)

This is because the VOLTMETER is reading the voltage across R2.

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FOIWATER
Gold Member
Yeah, what Duave is saying might be a good thing to remind yourself of often:

In the same way that I described the negative of the voltage source as IMPLIED as being connected to ground, it is similarly implied that the load voltage is taken with respect to ground. Of course as you said it is not necessarily! but unless otherwise noted that's what is being talked about.

In this example the voltages are taken across the resistors, though. From end to end

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jim hardy
Gold Member
Dearly Missed
You guys have stated it correctly, several different ways.

I think maybe it would be good to remind Maylis of this real basic fact:
When dealing with circuits-
Voltage is a potential difference, always between two points. That's why voltmeters have two leads.
That's why voltage is often demarked on drawings by an arrow indicating the two points between which you're measuring voltage.

The wikipedia sketches indeed failed to indicate between which two points Vout is taken, showing only one point. Where does the other voltmeter lead go ?

This is where hands-on labs are so important to an electronics curriculum.
Voltages are most often measured with one voltmeter lead attached to circuit common.
This is what the Wiki author assumed but did not show or state. He assumed it goes to circuit common , at the bottom.

If one did this as a lab exercise with real meter instead of imagining it, the firat question would have been :
"Why does my meter indicate zero regardless of the R1::R2 ratio? "
" Because you haven't hooked up the other meter lead."

Valid question, Maylis.
Buy yourself an inexpensive meter (\$10 at Walmart) and become skilled with it. It's real handy around the house.