Some insight here would be helpful because i am not getting the desired results..
According to this Book i am reading... Practical Electronics for Inventors...which is plagued with errors, which i did not find out until i purchased it and wrote in it.. but anyhow..
!0% Rule
Using there method to find the Value of R1 and R2 in a voltage divider circuit I first have to find the Bleeder current
which is the current through R2 and which is also 10% of the output current i want [i(load)]
so If i have a 5v power source and i want my output to be 3v at 25mA
i have to find 10% of the load/ output current i want
i2 (bleeder current ) = .1 x 25ma = 2.5 mA ... okay we will save this for later..
next it says to determine the load resistance, and that the bleeder resistance is 10 times the load resistance..
so being that i want 3v @ 25mA
R(load) = 3v / .025A = 120ohms ---> 120 x 10 = 1,200ohms
so R2(bleeder) = 1200 ohm
now to find the current i1 which is the one going through R1 i just add the bleeder current and the load current so : 2.5mA(i2) + 25mA (load/output Current) = 27.5 mA
so to find R1 i go back to ohms law now that i have the current for i1
R1 = (5Vin - 3Vout) / .027A(i1) == 74ohms
so R1 = 74ohm and R2 = 1200 ohms
and i should get an output of 3v and 25ma
but this is not the case 
my multimeter show .221 volts
obviously i am using slightly different resistor values but my output voltage is no where near what it should be according to the math, which i am sure i messed up somewhere along the lines
Just realizzed i posted this in the wrong section, i am so use to just coming to this section sorryy.
so If i have a 5v power source and i want my output to be 3v at 25mA
Voltage dividers make poor current sources. If the 3v load is a constant load of 25ma, just wire a single series resistor that drops 2 volts at 25ma = 80 ohms. If the 3 volt load current can vary, then you really should use a voltage regulator, like a zener diode or small 3 terminal voltage regulator chip.
Lefty
i mean i am not using this for a project or anything i just wanted to put the theory to work and see it for my self
From your results it would appear that you may have your 3volt/25ma load circuit in series with R2. R2 should be in parallel with the load, or to put it another way you get the 3volts by placing your load across R2.
Even on that basis the maths doesn't add up. With R2 = 1200 ohms together with R1 of 74 ohms and your circuit load of 120 ohms, assuming all are in series, the total series resistive load on the 5 volt supply is 1394ohms. This gives a load current of 5/1394 = 3.58mA. A current of 3.58mA through a 120 ohm load would develop 0.43 volts. Since you are only getting 0.221 volts, it appears your load resistance is actually lower than your expected 120 ohms.
As to the 10% rule, I think you may be misinterpreting the intent of the message. To get some degree of "stable" voltage out of any voltage divider network, the source impedance (R2) should be many times lower than the load impedance (typically 10% ?). So, if you are using R2 as your source, then R2 should be 10% of your load resistance. When 2 resistors (R1 and R2) are used in series to produce a voltage source (across R2), the load current should not be greater than 10% of the R2 current. In other words the current through R2 should be 10 times the current through the load. Using your load current of 25mA and a load resistance of 120 ohms, it follows that R2 should be 12 ohms with a current flow of 250mA. This means R1 is carrying 275mA and so needs to be only 2/.275 = 7.2 ohms
Now, I trust you see the problem. To get a reasonably stable 3 volts at 25mA, you are having to use a load current of 275mA from your 5 volt source. In other words, using resistors as potential dividers to provide power to an external circuit is a no-brainer. Potential dividers are excellent at providing "no-current" references but are useless for delivering power.