Add a resistor to invert a sine wave??

[Greg J.]

I must not understand what is going on at the negative side of the AC source vs. the ground in the attached diagram. Why is the signal at Node 2 ("n2") inverted?

 

[Alec_t]

The current through R2 and R3 is the same, but they are on opposite sides of the ground point.

 

[dorke]

A picture worth 1000 words...





The current in the loop is i.
Voltages on the resistors are:
VR1=i*R1
VR2=i*R2
VR3=i*R3
Voltages on the measurement points (vs.GND )are:
Vn1= VR2 = i*R2
Vn2= -VR3= -i*R3

It is not that "the signal is inverted",
It is just the way you measure and what is your reference point !
For instance, if you moved the reference point to be n2 itself ,
all voltages would measure the same "polarity".

Hope that helps

 

[Greg J.]

Indeed, thank you. Alec_t's answer reminded me that V1^- is not the reference point.