What is the Relationship Between Resistor Values in an Attenuator Circuit?

[profbuxton]

Homework Statement

Find R1 and R2

Relevant Equations

(R1+R2 +Ro) /R2 = 5 where Ro = SQRT(R1 x R2+(R1 x R1) / 4) = 60

Greetings, fellow travelers.
I gave been digging into some of my old (1970s) Electronic maths textbooks and for my amusement and time wasting refreshing some old knowledge. Have got through a fair but but have to keep coming back to this problem.
I do have the answers but I cannot regardless of how I manipulate the variables I can't seem to find the correct method obtain the answer. This problem is in part of the book related to solving quadratics but I can't see how this is related. Any broad hints at a solution would help me sleep better. Thanks for your assistance.

PS sorry about the format of the equations. Havent used the system to enter equations before. Maybe I will learn that too!

 

[BvU]

Greetings too.
What's the complete problem statement ? Any diagrams ?

 

[profbuxton]

sorry no diagrams with this problem.
The problem is to find the values of R1 and R2. I have tried manipulating the known answers to find the correct algorithm(big word that) but cannot find the correct method. I suspect it turns into some kind of quadratic but its been so long ago since I played with this stuff.

 

[Steve4Physics]

EDIT. Sorry - didn't spot that the value of R₀ is given. The reply (below) is therefore not relevant. See Post #6 instead.

There are three unknowns (R₀, R₁ and R₂) but only two equations. So you can’t solve for the values of R₁ and R₂.

Maybe you are meant to express R₁ and R₂ in terms of R₀. For example:
R₁ = 2.5 R₀ and R₂ = 0.2R₀
(not the answers, just some random made-up numbers for illustration).

If this is the case let R₁ = aR₀ and R₂ = bR₀. The problem is then to find a and b.

Substitute for R₁ (replace it by aR₀) and for R₂ (replace it by bR₀) in your equations.
You will get two equations containing only a and b. See if you can solve these.

 

[profbuxton]

Oh, sorry for the misunderstanding. The value for Ro is given by the equation
Ro = SQRT(R1 x R2+(R1 x R1) / 4) = 60 so in the first eqn. Ro actually equals 60.
sorry if it wasnt too clear . my bad. does that help?

 

[Steve4Physics]

Oh, sorry for the misunderstanding. The value for Ro is given by the equation
Ro = SQRT(R1 x R2+(R1 x R1) / 4) = 60 so in the first eqn. Ro actually equals 60.
sorry if it wasnt too clear . my bad. does that help?

Whoops – my mistake. Sorry. See if this gets you started...

Replace R₀ by 60 and you get two equations:
(R₁+ R₂ + 60) / R₂ = 5 (equation 1)
√(R₁R₂ + R₁²/4) = 60 (equation 2)

Rearrange equation 1 to make R₁ the subject.
Square both sides of equation 2 to get a new equation without the square root.
Can you take it from there?

If you need a reminder on how to solve quadratic equations, see here for example: https://www.bbc.co.uk/bitesize/guides/zwmyxfr/revision/1

 

[berkeman]

Oh, sorry for the misunderstanding. The value for Ro is given by the equation
Ro = SQRT(R1 x R2+(R1 x R1) / 4) = 60 so in the first eqn. Ro actually equals 60.
sorry if it wasnt too clear . my bad. does that help?

In LaTeX:
$$R_0 = \sqrt{R_1 R_2 + \frac{R_1^2}{4}}$$

See the LaTeX Guide link below the Edit window.