Simple Voltage Divider Algebra problem

[ECE101]

1: Vout = Vin.R2 / R2 + R1

2: (R2 + R1). Vout = Vin. R2

3: (-R2. Vin + R2) + R1. V out = 0

4: R2(-Vn +1) + R1.Vout = 0

5: R2 = (R1.Vout) / (Vin + 1)


How does one go from line 2, to line 3?

I end up with:

2: (R2 + R1). Vout = Vin. R2

3: (-R2. Vin) + R2. Vout + R1. Vout = 0

Could someone please explain?

 

[keith1200rs]

How does one go from line 2, to line 3?

By accidentally missing off the Vout from R2?

It would be easier to follow with the schematic.

Keith

 

[_Eduardo_]

1: Vout = Vin.R2 / R2 + R1 must be Vout = Vin.R2 /(R2 + R1)

2: (R2 + R1). Vout = Vin. R2 OK

3: (-R2. Vin + R2) + R1. Vout = 0 must be (-R2.Vin + R2.Vout) + R1.Vout = 0

4: R2(-Vn +1) + R1.Vout = 0

5: R2 = (R1.Vout) / (Vin + 1) error propagation.


If you need R2

by KCL (Vin - Vout)/R1 = Vout/R2
then R2 = R1.Vout/(Vin - Vout)