I have a simple question regarding a very simple formula

SN96 · 2006-01-10 21:25

I am embarrased to even ask, but I figure I would rather be laughed at and take it, then not learn something.

The question is very simple:

All I want to do is be able to figure out what voltage I will get when resistor(s) X is in series. I have looked on the net to find calculators that give watts, current, etc. but I would like to know voltages for starters.

Example circuit:

··············· 10k
5v <

/\/\/\

| GND = how many volts?

Thanks

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Mike

·

Paul Baker · 2006-01-10 21:45

Hmm Im slightly confused, the voltage drop across a single resistor will always be equal to the supply voltage. If you ware talking about multiple resistors and measuring the voltage between two of the resistors, the voltage will be equal to the supply voltage times the fraction of the resistance betwwn the tap point and ground divided by the total resistance. So for the simple example of two resistors configured as 5V---4.7kΩ---3.3kΩ---GND, measuring the voltage between the resistors wrt GND would be 5V x (3.3kΩ/8k&#937[noparse];)[/noparse] = 1.65V.

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·1+1=10

John R. · 2006-01-10 21:49

Mike;

In series with what? As shown in your post, you have 5V, there is no "formula".

If you're talking about this:

··················7k··········· 3k
5v <

^^^

^^^----| Gnd
························ |
························?v

Then you first have the fine the total current (I = 5/10,000 = 0.5 mA)

Then find the Voltage drop accross one of the resistors (E= 0.5mA * 3,000 = 1.5V)

The basic formula is E=I*R (Volts = Amps x Resistance)

John R.

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John R.

8 + 8 = 10

Beau Schwabe · 2006-01-10 22:02

With the example that you provided, you will always see 5V across the resistor.

Example #1:

+5V >---/\/\---< GND
         R

In this case you will always "see" 5V across R



Example #2:
+5V >---/\/\---o---/\/\---< GND
         R1    |     R2 
               o----------> Vout


Here,

Vout = Vsupply * (R2/(R1+R2))

OR

Vout = I * R2, where I = Vsupply / (R1+R2+...+Rn)

I personally like the second formula, because you can
easily determine the voltage across any resistor in the
chain once you determine the current.

Suppose R1=3K and R2=2K


I = +5V / 5K = 0.001 Amps = 1mA


Vacross_R1 = I * R1 = 3V
Vacross_R2 = I * R2 = 2V

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Beau Schwabe

IC Layout Engineer
Parallax, Inc.

SN96 · 2006-01-10 23:06

Thnaks, I was going to include a second example showing two resistors in series with a tap between the two but figured the first example would get me somewhere.

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Mike

Post Edited (SN96) : 1/10/2006 11:16:55 PM GMT

rockin_rick · 2006-01-11 22:36

To help with further searching, these circuits are typically known as voltage dividers. Try searching for that and you may find a calculator...

Rick

Guenther Daubach · 2006-01-11 22:56

Well, I don't think they teach this at law schools, but all this is handled by Ohm's law (plus Mr. Kirchhoffs comments)

.

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Greetings from Germany,

G

Beau Schwabe · 2006-01-12 02:21

G

I have a simple question regarding a very simple formula

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