Who's good at Math? Calculate Voltage Drop

[NickJuice]

Ok i have all the variables...and i know how to calculate for resistor needed

to reduce a 6v battery to 5v's with a 3.5ohm atomizer

its ((6v*3.5ohms)/5v)-3.5ohms= .7ohm resistor needed

NOW the question is....What if i have the the values of the Battery Voltage(6V) the Atomizer resistance(3.5ohms) and the Resistor Value(.75ohms)

How would that Equation look? Maybe its because its 1:20am but i just can't figure out the math

 

[Kewtsquirrel]

You need to find the voltage drop across the first resistor to see how much voltage is applied to the second (atomizer). Current in the circuit will be the same throughout, so first calculate that:

I=V/R(t) V=6v, R(total)= 4.25
I=1.411A

Now we calculate the voltage across each resistor, V=IR, so .9877v across the .75ohm resistor, and 4.93v across the atomizer.

You should really calculate using actual voltage tho, 2x 3v batteries are going to be sitting at 7.2v fully charged, and probably settle down to around 6.7v under load, then maybe 6.4v loaded for the voltage that will persist for the longest time while they discharge.

I calced it out to 5.2ohms as a good value that works for both 2x 3v and 2x 3.7v batteries, and thats what the atomizers I ordered were specced to, we'll see how they work here soon ^_^

 

[Nuck]

Ho ho...now I know who you are!!

BUSTED BUDDY!

 

[Belowme]

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