# voltage divider formula manipulation, (for the "bad at math")

Discussion in 'General Electronics Chat' started by ninjaman, May 17, 2016.

1. ### ninjaman Thread Starter Member

May 18, 2013
306
1
Hello,

and I am not good at math, or the rearranging of formulas.

Notes:

Though this “voltage divider formula” may be found in any number of electronics reference books, your students need to understand how to algebraically manipulate the given formulae to arrive at this one.

I am not too sure how to algebraically manipulate the given formulae. I have read the voltage divider text but understanding it is the problem.

Any help on this would be great!
Thanks
Simon

2. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
I'm not sure what you are referring to -- which formulae? Where is that Note quoted from? I'm lacking much context with which to work.

If you aren't good at algebra, the best approach is to work at getting reasonably proficient at algebra. Don't just try to learn "tricks" to work with specific formula in specific situations -- learn to understand and appreciate the math that is involved. It will take some time and effort, but the payoff will be enormous.

Do you understand where the basic voltage divider equation (where you have two resistors in series connected to an ideal source) comes from in the first place? That is as good a place to start as any.

cmartinez likes this.
3. ### crutschow Expert

Mar 14, 2008
13,475
3,362
You can manipulate simple formulas by multiplying or dividing both sides by the same factor.
For example, take Ohm's law, V = I*R.
To solve for R you divide both sides of the equation by I, giving V/I = (I*R)/I
Since (I*R)/I = R, you end up with V/I = R or, since you can swap values across an equal sign, R= V/I.

I leave it as an exercise for you to similarly solve for I.

Note that it may be simpler to first calculate the current through a voltage divider (especially if there's more than two resistors) than to calculate a voltage drop equation directly.
Thus you calculate the current through the series resistor string, using the total string resistance, and then calculate the voltage drop across each resistor from Ohm's law, V = I*R.
The voltage at any point in the the string is then just the sum of the voltage drops of the resistors below that point (towards common).

4. ### wayneh Expert

Sep 9, 2010
12,374
3,226
So true. If you work a lot of resistor divider example problems, you will probably learn the few manipulations that are typically required and get an intuitive feel for them after a while. But the better long run solution which will help you in so many other areas also, is to bite the bullet and improve your algebra skills. You won't regret it.

Apr 5, 2008
15,798
2,384
Hello,

Have a look at chapter 2.3 of the attached PDF.
(chapter 2 contains a lot of circuit theory).

Bertus

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6. ### DGElder Member

Apr 3, 2016
347
87
You might start here.

7. ### ninjaman Thread Starter Member

May 18, 2013
306
1
ER = Etotal * R
Rtotal

this is the formula that it talks about, it is quoted from question 3 in the worksheet mentioned in my first post. I know how to get V/R = I but the part about understanding how to get to the above formula is what is getting me. is this just rearranging. because i think it is understanding electronics as well. i have read a few things about voltage dividers. i have the art of electronics lab manual that i am trying to work through and a course i am trying to study (distance learning). the books and sites mention that you can look at a voltage divider and figure out, "intuitively" what the answer is going to be, without using all the calculations. i read about the idea of a load being in parallel with the output resistor and calculating using the parallel value and the "rule of thumb" of using a load that is a minimum of ten times the output resistor. and i understand the idea of an input having a high resistance. say an output of 10v into a load of 1 megohm would mean 10v cut up into 1 million bits and just one of those bits would be the current. so a high resistance would pull a small current. ......(i am correct in thinking this arn't....(sudden sinking feeling in pit of stomach)).

Apr 5, 2008
15,798
2,384
Hello,

Did you read the chapter I mentioned?
Here I have extracted the piece:

Bertus

Neil Groves 1 likes this.
9. ### GopherT AAC Fanatic!

Nov 23, 2012
6,303
4,019
Sit back and ignore the word "Algebra" for a minute.

Think about two resistors in series. The total resistance is R1 (connected from ground to the other resistor (R2)), and R2 (connected from R1 to Positive battery).

Now, the total resistance is just R1 + R2 . Easy.

The current in this "series" circuit = battery voltage / Total resistance

Good?

Now, the voltage at the node between R1 and R2 is simply what?

Current * R1.

Don't over complicate things.

Always work from Ground as your reference. That means, calculating based on R2 would be more difficult because we would have to subtract from the supply voltage to get the voltage between R1 and R2.

Ohm's Law works. You don't have to call it Algebra, just common sense can get your answer.

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10. ### MrAl Distinguished Member

Jun 17, 2014
2,551
515

Hi,

I believe the question you are referring to is question 15.

Back in the 1980's i was unsatisfied with 'regular' circuit analysis to some degree so i developed another way of thinking about a voltage divider. This alternate view of a voltage divider still requires some circuit analysis concepts, but only requires understanding Ohm's law in a very simple way and the concept of equality.

First, it's a good idea to learn algebra better because it's more widely applicable and you can solve an awful lot of problems using just algebra.

To get another view of the voltage divider to begin with, we know that in the most basic divider there are two resistors in series, and the first law we use is:
"Circuit elements in series all have the same current".
What that means is simply that each resistor MUST have the very same current level. If R1 has 1 amp, the R2 must have 1 amp too.

Next, we know Ohm's Law:
E=I*R

This tells us the voltage ACROSS each resistor, knowing the current I and each resistor value.

The alternate view is to think about how nature itself might be trying to settle on a voltage at the center connection of the two resistors. Because of other physical constraints, it may start out with a voltage of one half of the supply voltage, even if the resistors are not equal in value. This might happen because the resistors are the same physical size (an assumption for this problem of course).

For a simple example, say we have a resistor R1 on top of 1 ohm and a resistor R2 on bottom of 2 ohms, and a power supply voltage of 3 volts. What is the output voltage Vout of the divider?

We start by calling Vout=1.5 volts, because that is one half of the total supply voltage. Next we test this to see if it is right, and if not we modify it.

Using Ohm's Law for R1 we get:
iR1=1.5/1=1.5 amps

then for R2:
iR2=1.5/2=0.75 amps.

Since iR1 is not equal to iR2, we know this isnt right yet. To get these two currents to be equal we must have a different voltage for Vout. Since iR2 is lower then iR1, if we raise the voltage a little then the two currents will be more equal. Raising Vout by 0.1 volt we now say that Vout=1.6v instead of 1.5v.

Doing the equations for iR1 and iR2 again we get:
iR1=1.4/1=1.4 amps
iR2=1.6/2=0.8 amps.

We see we already made progress, because the difference between the two currents is less now then before.

We'll do this one more time, increasing Vout by 0.2v just to see what will happen, we we get Vout=1.8v and:
iR1=1.2/1=1.2 amps
iR2=1.8/2=0.9 amps.

Now we see we are much closer to equality. If we raise Vout again by another 0.1v we get Vout=1.9v and:
iR1=1.1/1=1.1 amps
iR2=1.9/2=0.95 amps

and now we are much closer than when we started. Raising Vout by one more 0.1v gives us Vout=2v and:
iR1=1/1=1 amp
iR2=2/2=1 amp

and now both currents are the same, so we have the right value for Vout.

This procedure should tell you at least two things:
1. Nature is a lot more complicated than circuit analysis
2. It's much easier to use algebra than a hunt and peck trial and error method.

Moral of the story: Although this actually is an alternate method to find the right voltage (and there are more mathematical concepts that can be applied that reduces the trial and error) using algebra gets us to the right result with just one single try, so learning algebra is a very good idea as it will help gain knowledge in other areas much faster. Imagine if we had to do trial and error with every circuit.

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11. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
One thing that will help communication immensely will be to typeset equations so that they are easily readable. Trying to do it in a sentence they way you did almost always leads to confusion. On the forum, that is usually best done using LaTeX -- look at the Stickies in a couple of forums and you will find a tutorial on it.

What you meant to write:

$
E_R \; = \; E_{total} $$\frac{R}{R_{total}}$$
$

Everything in this part is just applying Ohm's Law

$
E \; = \; I \times R
$

It's critical to keep in mind that Ohm's Law relates a particular resistance to the voltage across that SAME resistance and the current through that SAME resistance.

The total current in the series circuit is governed by

$
E_{total} \; = \; I_{total} \times R_{total}
$

So this is the total voltage applied across the series combination of resistors, the current flowing through each of these series elements, and the total resistance of all of these series elements.

Dividing both sides by R_total gets you

$
\frac{E_{total}}{R_{total}} \; = \; I_{total}
$

Now pick one of the resistors and you know that Ohm's Law applies to it, too.

$
E_{x} \; = \; I_{x} \times R_{x}
$

We can replace I_x with ANY expression that is equivalent to I_x.

The current in Rx is the same as the total current (because the resistors are in series), so we know that

$
I_{total} \; = \; I_{x}
$

So, replacing I_x with the I_total, and then an expression that is equal to I_total, we get

$
E_{x} \; = \; I_{x} \times R_{x}
E_{x} \; = \; I_{total} \times R_{x}
E_{x} \; = \; \frac{E_{total}}{R_{total}} \times R_{x}
E_{x} \; = \; E_{total} \times \frac{R_{x}}{R_{total}}
$

12. ### Tonyr1084 Active Member

Sep 24, 2015
628
109
In the drawing below you see a 20 volt source and four resistors (R1 through R4). All resistors are in series. On the left side you see the resistor values listed. Add them together and you see a total resistance of 8.9KΩ.

Everyone has told you that I (amperage) is equal to E (voltage) over R (ohms or resistance). Swap that around from I=E/R to E/R=I (it's the same thing either way) and substitute the values. I (unknown) is equal to E (20 volts) over (or divided by) R total (8.9KΩ). when you divide 20 by 8.9 you get 2.25. THAT is how much current is flowing through all resistors at the same time. What passes through one passes through all for a total of 2.25 mA (milli-amps). Since 8.9K is the same as 8900 you can divide 20 by 8900 and you'll get 0.00225 amps. That's the same as 2.25 milli-amps. You now have two numbers to figure the voltage drop across ANY resistor. E=I*R, so if you KNOW you have 2.25 mA and 1KΩ then 2.25 x 1 = 2.25 (volts dropped across R1). You can do the same for each resistor value. 2.25 x 4.7 = 10.58 (volts dropped across R2).

Test it out. What do you get when you multiply 2.25 x 2.2? Since R4 is the same value as R1 - the answer should be obvious. The current is always the same through the whole (series) circuit. At any point you can calculate the voltage to ground by subtracting the voltage drop (or drops) of all resistance above the point you're testing.

In the drawing you see the volt drop across each resistor AND the voltage from a point (between R1 & R2 for example) to ground. This is the method I learned and have used for many years. This works nice in strictly series circuits. And I'm certain someone else may have a method they like better. I'm not saying mine is the best but it works for me.

In the drawing you see I've totaled up all the voltage drops for a total of 20.03. But I started with 20 volts. Did I gain something? No. I simply rounded my voltages to the nearest 100th. But 3 100ths is close enough.

Last edited: May 17, 2016
13. ### hp1729 Well-Known Member

Nov 23, 2015
2,086
232
Has all this helped you? Do you need another approach?

14. ### hp1729 Well-Known Member

Nov 23, 2015
2,086
232
Do you have Microsoft Excel? Can you use a spreadsheet if all you have to do is fill in data into a cell and let the spreadsheet do the calculating for you?
I can't post an Excel file here in the forum but I can email it to you. We need to get in touch through a PM (Personal Message) to exchange e-mails.

Attached is a PDF of what the excel files look like.

I have a lot of electronics calculations done up in Excel if anybody is interested. some less friendly than others.

• ###### Voltage Dividers.pdf
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15. ### Tonyr1084 Active Member

Sep 24, 2015
628
109
Nice spreadsheet. Good for just two resistors. Here's mine (good for up to 10 resistors) Notice all unused resistors are at a value of zero. They contribute nothing to the formula except that they hold place for total resistance and current calculations. Simply enter the voltage supplied and the used number of resistors and their values. In this case I wanted to modify a circuit I copied from AAC but their power source was (I think) 34 VDC. Notice the voltage I used is less. That meant the target 2.5 volt bias (R3) would have been lower. So I modified R2 until I found the needed value of 2.5 volts.

This sheet (and yes, this is PDF) can calculate voltage drops of up to 10 resistors. Beneath each resistor is the associated volt drop across that resistor. Beneath that is the remaining voltage to ground. I have this spread sheet available in my DropBox folder and I can provide a link to anyone who wants to play with it.

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