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Suppose I have the following circuit:

enter image description here

Are resistors R1 and R2 in parallel, assuming that through resistor R1 flows current because this circuit is connected as a feedback to another circuit? And if so, why?

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No, R1 and R2 are not in parallel unless the load was 0 Ω (so vout = 0).

Assuming the load is non-zero, but it is not a simple resistive load that can be separately measured, you can get an equivalent resistance by measuring the current through the load and the voltage across it.

$$R_{load} = \frac{vout}{I_{load}}$$

Then, the resistance from the junction of R1 and R2 to ground would be:

$$R_{parallel} = \frac{R1 + R2 + R_{load}}{R1 \times (R2 + R_{load})}$$

Note that if you plug in 0 for \$R_{load}\$ in the above equation, you get the formula for just R1 and R2 in parallel, as stated in the first paragraph.

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R2 and R1 are not in parallel. R2 and the sum of R1 plus whatever resistance is placed across vout are in parallel:

R2 || (R1 + Rvout)

The calculation of the resistance at vout may be a non-trivial calculation, but in every case, the voltage across R2 will be the same as the voltage across R1 plus vout:

VR2 = VR1 + vout

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  • \$\begingroup\$ and if we assume that RVout is 0 why R2 is in parallel with R1? \$\endgroup\$ – John Doe May 25 '15 at 16:50
  • \$\begingroup\$ @JohnDoe If Vout is 0 it is like having a wire connecting the + and - terminals, this makes R1 and R2 connected in parallel. \$\endgroup\$ – Lorenzo Donati May 25 '15 at 17:30
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No, they are not in parallel. Think about it this way: In order for R1 to be in parallel with R2, its right-side would need to be connected directly to ground. However, if it was, then there would be no voltage at vout--It would be measuring between ground and ground (0 volts difference).

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To the base of the transistor (only) they look like they're in parallel.

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