Engineering Intro EE - Voltage Divider Design circuit

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The discussion focuses on designing a voltage divider circuit using specific resistor values. The primary equations utilized include the voltage output ratio and Thevenin resistance, with constraints set for R_th between 10 and 30. Initial calculations for R_1 yielded values of 56k, 68k, and 82k, leading to corresponding R_2 values of 14k, 17k, and 20k. A correction was noted regarding the resistance range, clarifying that R_th should actually be between 10,000 and 30,000 ohms. The conversation emphasizes the importance of considering all available resistor values in the E12 series for accurate results.
krtica
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The link is an image with the question and circuit representation:
http://i49.tinypic.com/2mmwg8l.png


Here are the equations I used:

V_out/V_in = R_2/(R_1 + R_2) {using the node method}

R_thevenin (R_th) = R_2*R_1/(R_1+R_2)

And given the conditions =>

10 < R_th < 30

Manipulating this inequality I found that 50 < R_1 < 150, which left me with three possible values for R_1 (56, 68, 82).

Using the first equation, I put in the numbers, which gave me 0.2=R_2/(R_2+56) and solved for R_2 with every value I put in for R_1 (56, 68, 82).

R_2 values were 14, 17, and 20, respectively, which came to equate to 15, 18, and 22 when compared to the original set of resistors and the 10% condition.

I tried putting in the values of these resistors I used, because all gave me the right ratio of voltages, but they're incorrect.


Pardon how nuanced I may seem, it's been too long since I've done any physics..
 
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krtica said:
Here are the equations I used:

V_out/V_in = R_2/(R_1 + R_2) {using the node method}

R_thevenin (R_th) = R_2*R_1/(R_1+R_2)
I think you are on the right track.
And given the conditions =>

10 < R_th < 30
This should be, 10,000 ≤ R_th ≤ 30,000
Manipulating this inequality I found that 50 < R_1 < 150, which left me with three possible values for R_1 (56, 68, 82).
Aren't you missing some candidates here?
R₁ ∈ {56k,68k,82k,100k,120k,150k}
 
Wouldn't it just be the three since your given all the resistances available in the set E12?
 
120kΩ = 12 x 104
 
Right! Thanks.
 
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