Electronic Circuits and Projects Forum

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: I²R
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.

So just add them up:

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.