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Re-arranging curved graph to get a straight line Watch

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    Can somebody tell me if I've re-arranged this formula correctly, I am trying to turn it into a straight line

    VOUT = R2 / R1 + R2 * VIN

    1/VOUT = R1 + R2 / R2 * 1/VIN
    1/VOUT =R1/R2 + R2 / R2 * 1/VIN

    1/VOUT =R1/R2 + 1* 1/VIN

    1/VOUT =R1/R2 + 1/VIN


    so
    y = 1/VOUT
    m = R1
    x = 1/R2
    c = 1/VIN

    my only problem is that my y-intercept does not become 1/vin. So i am doubting whether I have rearranged the formula correctly. Can somebody please clear this up for me.

    Thanks in advance

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    (Original post by HK-97)
    Can somebody tell me if I've re-arranged this formula correctly, I am trying to turn it into a straight line

    VOUT = R2 / R1 + R2 * VIN

    1/VOUT = R1 + R2 / R2 * 1/VIN
    1/VOUT =R1/R2 + R2 / R2 * 1/VIN

    1/VOUT =R1/R2 + 1* 1/VIN

    1/VOUT =R1/R2 + 1/VIN


    so
    y = 1/VOUT
    m = R1
    x = 1/R2
    c = 1/VIN

    my only problem is that my y-intercept does not become 1/vin. So i am doubting whether I have rearranged the formula correctly. Can somebody please clear this up for me.

    Thanks in advance

    What are you trying to do?

    Vout will follow whatever voltage is input as a ratio between R1 and R2.

    It's already a straight line as R1 and R2 are linear components.

    R1 an R2 simply form a potential divider.

    In this case y=f(x) where y is the output voltage Vout; x is the input voltage Vin; and f(x) is the input voltage * the ratio R1/(R1+R2).

    The gradient is therefore given by m=R1/(R1+R2) and starts at the origin since 0*R1/(R1+R2) = 0

    In this case c=0

    If you are looking for the impedance of reactive components, that is a different matter.
 
 
 
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