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Re-arranging curved graph to get a straight line watch

1. 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.

2. (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.

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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