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Thread: anyone know how to solve this? (my elect's class) quick ?

  1. #1
    FLAC guitar333's Avatar
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    anyone know how to solve this? (my elect's class) quick ?

    I am having a hard time with this one: (not meant to be a long thread, just need solution)
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  2. #2
    Newbie PSUSuperSenior's Avatar
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    1.5 kohms, taken across the load terminals. Thevenin voltage is (Va / 8) if you need that too. If you want to know how, just ask

  3. #3
    FLAC guitar333's Avatar
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    i know you remove the load and calculate resistance but thats where I get lost, if you can just tell me how you got 1.5k then I will be set.

    (sorry I have no instructor to ask physical ?'s, online class)
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  4. #4
    Variable Bitrate SilentAdmirer's Avatar
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    To calculate the the total resistance also replace all the voltage source by short circuit and current source with open circuits.

    So if u start from left most side
    R1 in parallel with R2 = R1-2 = 1K
    R1-2 in series with R3 = R1-3 = 2K
    R1-3 in parallel with R4 = R1-4 = 1K
    R1-4 in series with R5 = R1-5 = 2K
    R1-5 in Parallel with R6 = R1-6 = 1K
    R1-6 in series with R7 = Rth = 1.5K

  5. #5
    FLAC guitar333's Avatar
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    thanks, makes sence now
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  6. #6
    FLAC guitar333's Avatar
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    hehe, one more ?

    sorry, I wont clutter the boards up anymore after this

    could one of you guys just show steps, I have looked at examples in books and site and what not but they deal with voltage dividers, so one example will help me VERY much

    APPRECIATE IT!
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  7. #7
    Constant Bitrate
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    I get 0.035mA for I Load. (not one of the answers given). got a thevenin equivalent circuit (Vth = 0.475V, Rth = 3.6kOhms) and used that to calc the load current. Must have stuffed up one of the thevenins.

  8. #8
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    wow

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