misterT is correct. If you really want to go down the resistor path then do the above.
But i still maintain that if you then put a significant load, lets say .5A on the output then you'll totaly unballance your resistor divider. You can only really use the resistor solution if you know your desired load current and take this into account during the calcuation.
There is this one time I needed to calculate a voltage divider to drive a LED. But that was a special case for bicolor LED (and I wouldn't call it an application of voltage divider):
What are you trying to drive with the voltage divider? What is the practical application you need a solution for? We can't give you any rules to a problem that does not exist. If you don't have a practical application, make something up as an example.
misterT, not sure why you needed R1 in that circuit. The led voltage drop would of regulated the voltage for you, all you needed to do was limit the current - this is why R1 is usualy called a current limiting resistor.
In any case your final comments are right, we need more info to solve this question.
ok now i understand , if i connect some circuit in parallel to R1 which requires 2V to operate, then the current division will take place , which will change the voltage drop from 2V to something less, and so resistor value of R1 would have to be changed to get 2V back !
yes thats what i said
so Z= √ 10k^2 + 31400^2
then Z= 32953 ohms
then current I= V/Z .... 220/32953 = .00667 Amps
or 6 mA but LT spice is giving me 40mA peak to peak which is 30mA RMS. what is going wrong ?
yes thats what i said
so Z= √ 10k^2 + 31400^2
then Z= 32953 ohms
then current I= V/Z .... 220/32953 = .00667 Amps
or 6 mA but LT spice is giving me 40mA peak to peak which is 30mA RMS. what is going wrong ?