# Are these resistors in parallel?

Suppose I have the following circuit:

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?

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.

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

• and if we assume that RVout is 0 why R2 is in parallel with R1? – John Doe May 25 '15 at 16:50
• @JohnDoe If Vout is 0 it is like having a wire connecting the + and - terminals, this makes R1 and R2 connected in parallel. – Lorenzo Donati May 25 '15 at 17:30

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).

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