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.
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.

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