Resistor Math Problem: Finding Equivalent Resistance in Series and Parallel

[Ithryndil]

Homework Statement

Two resistors connected in series have an equivalent resistance of 744.7 Ω. When they are connected in parallel, their equivalent resistance is 130.4 Ω. Find the resistance of each resistor.
Ω (small resistance)
Ω (large resistance)

Homework Equations

Req = R1 + R2 for Series.
1/Req = 1/R1 + 1/R2 for Parallel.

The Attempt at a Solution

Ok, so I began by solving one for R1 and plugging it into the other...

If I solve the first one I get R1 = Req - R2. Let's call Req, Reqs for Equivalent Resistance in series. Let's call Reqp the Equivalent Resistance in parallel.

So I plug into the other one:

1/Reqp = 1/ (Reqs - R2) + 1/R2

Messing around I am able to get:
R2Reqs - R2^2 = Reqs*Reqp However when I graph if there is no sign change... I've done it a few times so I don't think it's my algebra.

 

[nasu]

Try to solve the quadratic equation.
It seems OK to me. Actually I've solved it and it has two real solutions.
You did OK.
Good luck.

 

[lax1113]

Hey,
yeah when i used quadratic eq. it seemed to work out fine.

I learned resistors with the equation (R1*R2)/(R1+R2) which is the same thing simplified, but i always find htis one easier.

Anyway, it seems like you just about had it, just needed to finish it off with the eq and you would have it.

 

[LowlyPion]

Homework Statement

Two resistors connected in series have an equivalent resistance of 744.7 Ω. When they are connected in parallel, their equivalent resistance is 130.4 Ω. Find the resistance of each resistor.
Ω (small resistance)
Ω (large resistance)

Homework Equations

Req = R1 + R2 for Series.
1/Req = 1/R1 + 1/R2 for Parallel.

The Attempt at a Solution

Ok, so I began by solving one for R1 and plugging it into the other...

If I solve the first one I get R1 = Req - R2. Let's call Req, Reqs for Equivalent Resistance in series. Let's call Reqp the Equivalent Resistance in parallel.

So I plug into the other one:

1/Reqp = 1/ (Reqs - R2) + 1/R2

Messing around I am able to get:
R2Reqs - R2^2 = Reqs*Reqp However when I graph if there is no sign change... I've done it a few times so I don't think it's my algebra.

Your equation is ok, but you need to solve the quadratic, and since the 4ac term is positive you should have a real result.

 

[Ithryndil]

Bah, it must have been a stupid graphic error on my calculator. I knew I needed to solve the quadratic and must have just goofed when inputting on my calculator to graph and then find where it intersects y = 0. Thanks.

 

[lax1113]

I'm guessing that by the way you are talking, you have either a TI-83, or maybe a more advanced. If you are only using an 83 or 84, which is what most people have, there's a very easy quadratic program that you can put in there, (like write yourself) so you don't even need a cord to hook it up to your PC. I made one for mine, and it really comes in handy, because not only its quicker, but then it lessens any stupid mistakes you might make, (like accidently not putting a negative, or having to deal with a window issue for graphing).

So this isn't the exact same as the one i made, but it definately works, just might be one step more than needed. http://www.math.montana.edu/math105/QuadProg.html

Good luck!