Thevenin equivalent question

[Addez]

If I have 2 resistors R1 and R2 in series and I want to create a thevenin circuit from the voltage across R2 and ground.
Then Uth = Vcc * R2/(R1+R2) = lets say 7V
so theres 7V potential in the point between the two resistors.

Then I calculate Rth and put it as a resistor in between the Vth source and the point between R1 and R2.
Rth will undoubtedly cause a voltage drop, resulting in me NOT getting 7V between R1 and R2.


How can I then call it an "equivalent" circuit when it's not even giving correct voltage?

 

[DWard]

You're doing it wrong. The Thevenin equivalent is the Thevenin voltage in series with the Thevenin resistance. The thevenin resistance in this case is R1 in parallel with R2. If a resistance R3 is connected between the "Thevenin" node and ground (i.e. in parallel with R2), you can find the node voltage as in your example (replacing R2 with R3 in parallel with R2), or you can use the Thevenin equivalent - Vnode = Vth * R3/(Rth+R3). They are the same.

 

[D.A.(Tony)Stewart]

If you have Vcc=12V with Rs=0 and you want Uth=7V with Rs=x such that R_load >> Rs
Then R2/(R1+R2) = 7/12
where (R1+R2)/R2 = R1/R2+1 = 12/7
or R1/R2 = 5/7 .........(1)
If x = say 1000 and 1/1000= 1/R1+ 1/R2 ........2
Now solve for R1 and R2 using .......1 & 2

 

[DWard]

If I understood your post correctly, this has gone from a Thevenin equivalent question to an algebra question. To solve for R1 and R2, let's express them as some multiple of R (i.e. R1 = 5R and R2 = 7R), which is a kind of restatement of (1), but putting the unknown as a multiple of R. So, restating (2) in terms of R, 1/1000 = 1/(5R) + 1/(7R).
First: Multiply both sides by 35R
35R/1000 = 7+5 = 12
Next, Multiply both sides by 1000
35R = 12000
Now, divided both sides by 35
R = 12000/35 = apx. 342.857, or wherever you deem the need to round.
So R1 = 5R = 1714.286, and R2 = 7R = 2400.