Smaller and larger resistance in circuit

[nothingatall]

Homework Statement

When resistors 1 and 2 are connected in series, the equivalent resistance is 14.7 Ω. When they are connected in parallel, the equivalent resistance is 2.56 Ω. What are (a) the smaller resistance and (b) the larger resistance of these two resistors?

Homework Equations

1/Req=1/R1+1/R2 parallel
R1+R2+Req series

The Attempt at a Solution

My t.a. said that there would be a substitution and then end up with a quadratic equation. So far i have: R1+R2=14.7ohm
1/R1+1/R2=1/2.56ohm
R1(R2)=2.56(R1+R2) => R1(R2)=37.632ohm
2.56R1+2.56R2-37.632=0 <= my incorrect quadratic which i don't know how to correct.

How can i develop a quadratic from my info? Thanks!

 

[collinsmark]

Hello nothingatall,

The Attempt at a Solution

My t.a. said that there would be a substitution and then end up with a quadratic equation. So far i have: R1+R2=14.7ohm
1/R1+1/R2=1/2.56ohm
R1(R2)=2.56(R1+R2) => R1(R2)=37.632ohm

Okay, I follow you up to here. So far things seem okay. But you've missed one of the substitutions. what you'd like to do is get an equation that is completely in terms of either R₁ or R₂. (See more below.)

2.56R1+2.56R2-37.632=0 <= my incorrect quadratic which i don't know how to correct.

Nothing particularly bad, but you're failing to make a substitution. You know that

R₁+R₂ = 14.7 Ohm.

That means you also know that

R₂ = 14.7 Ohm - R₁.

Using that knowledge, now go back to on of your earlier equations (involving the equivalent parallel resistance) and try to put everything in terms of R₁. Then solve for R₁. Once you've found R1, you can use the above equation to get R₂.