# Electronic Circuits and Projects Forum

1. ## Determine Voltage Drop Across Resistors

I learned this from http://www.bcae1.com/resistrs.htm
I know some true nooBs would like this info so i decided to post it again

How to calculate voltage drop across series resistors.

Step 1: Determine Current of total resistors across the series.
(I = Current/V=Voltage/R(t)=Resistance Total)
I = V / R(t)
I = 9 / (2k + 5k + 10K)
I = .53 mA

Step 2: Now that we have the Current across the resistors we can calculate the voltage across EACH resistor.
I = Current
Vx=Voltage (x=Current Value across X Resistor)
Rx=Resistor value(x=resistor#)

V = I x Rx

V1 = .53 mA x 2k
V1 = 1.06V

V2 = .53 mA x 5k
V2 = 2.65

V3 = .53 mA x 10k
V3 = 5.3V

V1 + V2 + V3 = 9.01 (voltage is 9V only 9.01 because i rounded up from .529 to .53) At .529 you can see barely a difference:

V1=1.058V
V2=2.645V
V3=5.290V
V Total= 8.993V

Also If the resistors are all the same value its easier.

From: IČR
http://www.electro-tech-online.com/m...witches-2.html

To determine the voltage applied by each keypress, you simply add up the voltage drops connected to that key.

I = V / R(t)
I = 9 / (5k + 5k + 5K)
I = .6 mA

THEN:

V = I x Rx

V1 = .6 mA x 5k
V1 = 3V
so since all Resistors are same value:
V2 = 3V
V3 = 3V

Also To get different values for like switches just use gnd as a start point and the end of each resistor and a end point. This way you get 3 switches from this each having different voltages

SW1 = 3V
SW2 = 6V
SW3 = 9V

So I hope someone can learn something from this.

This Voltage Drop Across Resistors article is now in the Theory section.

2. Thank for posting this, Im quite beside myself with being dejected for being to stupid .
Im trying to figure out how to figure the voltage drop over to resistors.
The example Im using is a DC circuit
V=25
R1=3.3
R2=2.5
I =6.3
I told ( according to an quiz) that the voltage drop @ R1 is described as 21V
and R2 as 14
I have know idea how to come up with those values.
Any insight would be greatly appreciated.

3. ## It is wrong.......

Originally Posted by FITNAH
Thank for posting this, Im quite beside myself with being dejected for being to stupid .
Im trying to figure out how to figure the voltage drop over to resistors.
The example Im using is a DC circuit
V=25
R1=3.3
R2=2.5
I =6.3
I told ( according to an quiz) that the voltage drop @ R1 is described as 21V
and R2 as 14
I have know idea how to come up with those values.
Any insight would be greatly appreciated.
Ohms law says that E/R=I, so I should be 4.31A
So if E= IxR what do you get?

4. 34.65a.
I get 6.36 for I ,dividing the combined resistance into V.
which then in turn gives me a voltage drop of 20.79 =R1 and 15.75 =R2

5. Originally Posted by FITNAH
34.65a.
I get 6.36 for I ,dividing the combined resistance into V.
which then in turn gives me a voltage drop of 20.79 =R1 and 15.75 =R2
How can 25v/5.8 ohm be anything but 4.31A?

6. Sorry typo, the model I have here is V =35
4.54 is of course correct, sorry about being sloppy.

7. I already feel like quit the idiot over not knowing how to find this answer and now flubbing the question hasn't helped,
In any case their must be a formula for finding the voltage drop over two or more resistors of different values

8. Originally Posted by FITNAH
I already feel like quit the idiot over not knowing how to find this answer and now flubbing the question hasn't helped,
In any case their must be a formula for finding the voltage drop over two or more resistors of different values
Ohms law and a little thought!.

9. Thanks, the voltage drop across the resistors must add up to the voltage correct?

10. For a start draw the circuit and post it here!.

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