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# Re-arranging curved graph to get a straight line Watch

1. I have already re-arranged the potential divider formula from Vout=R2/(R1+R2 )*Vin to 1/Vout =(R1+R2/R2) * 1/Vin

My question to you is how I can rearrange this further to get
1/Vout =( R1/Vin)*(1/R2) + 1/Vin.

I am doing this so I can get it in the form y=mx+c
2. (Original post by HK-97)
I have already re-arranged the potential divider formula from Vout=R2/(R1+R2 )*Vin to 1/Vout =(R1+R2/R2) * 1/Vin

My question to you is how I can rearrange this further to get
1/Vout =( R1/Vin)*(1/R2) + 1/Vin.

I am doing this so I can get it in the form y=mx+c
EDIT.

The original equation is already in the form y=mx

i.e. y= Vout; m = R2/(R1 + R2); x= Vin

So what's the problem?

Hint, simply by inspection, the circuit is a potential divider and Vout is across R2.

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Updated: April 16, 2013
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