thanks for the insight guys...how about a little ohm's law help then...i have a 100uf 2watt resistor...while the board is powered up i get 25v on one leg and about 5v on the other...seems like alot of voltage drop to me...thanks in advance for the calculation
Well, that's not really possible to do with the information you have given. You'd have to know the TOTAL RESISTANCE of the entire circuit that part is located in.
Check this out:
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
Resistors that are in parallel will all have the same voltage drop...