How Is Thevenin Resistance Calculated in R1 Parallel to R2 Configuration?

[theycallmevirgo]

TL;DR

How is thevenin equivalent calculated for this network

The attached image (Horowitz & Hill Student Manual for 2nd ed) gives thevenin resistance and voltage for the network R_1+(R_2||R_L) as

R_th=R_1||R_2

E_th=E_in(R1/(R_1+R_2)) E_th is self-evident (voltage divider) but how in the world they get R_th i beyond me. If there is no load, isn't R_1 in series with R_2 no matter what?

Many thanks in advance

Joe

 

[Baluncore]

If there is no load, isn't R_1 in series with R_2 no matter what?

If you consider the ground and Vin to both have zero impedance, then the impedance of the R1-R2 junction is seen as the two resistors in parallel.

 

[theycallmevirgo]

If you consider the ground and Vin to both have zero impedance, then the impedance of the R1-R2 junction is seen as the two resistors in parallel.

So if I understand you correctly, wrt resistance gnd and v are "the same node" and therefore R_t is "measured across" R1-R2 junction and ground/v "node"?

 

[Baluncore]

Yes.
An ideal voltage source has a zero impedance, so ground Vin when calculating the thevenin resistance of the R1-R2 junction node.

If the supply was instead an ideal current source Iin, you would disconnect it as it would have an infinite impedance, then Rth would be only R2.