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bleeders and what I am doing....

Discussion in 'General Electronics Discussion' started by tedstruk, Apr 27, 2017.

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  1. tedstruk


    Jan 7, 2012
    I found this great resource on setting resistance and adjusting load, but I closed my laptop and lost the resource.
    Setting the resistance I understand fully.

    10% RULE-
    1. SET RESISTANCE where the amperage is 10% of the LOAD
    [current bleed]
    where I is amps

    IBleed =
    I =(0.10)(9.1a)=.91ma ( remark from dummystruk "(0.10 is 10% and 9.1a is the load" )

    [resistor bleed]
    Where E is volts and
    Where R is resistance
    RBleed = E / I = R
    R = 3v / .00091a = 3297

    Select a resistor near the RBleed value - R1 = 3300
    (dummystruk OK to here)
    Then it says to ---

    select a resistor to maintain the voltage.

    I1 = I2 + ILoad = [.91ma + 9.1ma] = 10ma = .010a

    R2(10v-3v) / .010a = 700 ohms

    I did it in my head the first time through... everything was great... then I did it again later, and I am missing the 700ohms every time....

    What I am doing wrong... I don't know what I2 is...
    Is I1 = 9.1ma and I2 = .91ma?
    R2 is my new resistance that maintains voltage, 10v-3v is 7/.010a = 700 the New resistor that maintains voltage is R2 = 700ohms

    wow it worked! I don't believe it!

    I wonder what I was doing wrong!!!

    So what this 10% RULE is about is creating a circuit that works under load and does not use up the voltage?
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    What are you trying to do? I really can't understand what you're up to.
  3. AnalogKid


    Jun 10, 2015
    1. I don't see a question in there.
    2. Open the browser history (Ctrl-H) to see past websites.

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