Resistance Problem

[RustyLeeNorris]

If two resistors R1 and R2 are in parallel then 1/R= 1/R1+1/R2 where R is the total resistance. Suppose the total desired resistance is 9 ohms. The first resistor needs to have a resistance 80 ohms greater than the second. What should their resistances be?

 

[RustyLeeNorris]

Resistance

I am stuck at trying to simplify 1/9= 1/r1 + 1/r2. I am lost , The forum post says to show work but I don't know how to implement the ohms in the problem, sorry.

 

[stapel]

If two resistors R1 and R2 are in parallel then 1/R= 1/R1+1/R2 where R is the total resistance. Suppose the total desired resistance is 9 ohms. The first resistor needs to have a resistance 80 ohms greater than the second. What should their resistances be?

The first resistor is defined in terms of the second. So what variable did you choose for the resistance of the second? What expression then stands for the resistance of the first? When you plugged these, along with the given value, into the formula, what did you get? How far have you gotten in solving this rational equation?

 

[RustyLeeNorris]

Sorry, I am just totally lost and I don't know where to put the "80" from the variable. I tried to get R,R1 and r2 by themselves and got R1*R2/R2+R1 (for r), RR2/(R2-R) for R1.

 

[lookagain]

Sorry, I am just totally lost and I don't know where to put the "80" from the variable.
I tried to get R,R1 and r2 by themselves and got R1*R2/R2+R1 (for r), RR2/(R2-R) for R1.

Your variables are R, R1, and R2. You are using upper case. Stay in upper case.

In one of your posts, you wrote this: "1/9= 1/r1 + 1/r2."


Don't write that, but write this: 1/9 = 1/(R1) + 1/(R2) \(\displaystyle \ \ \) *


Just to be clear, let R1 be the first resistor and let R2 be the second resistor.


You stated that you don't know where to put the "80."


Look at this statement from the problem:


"The first resistor needs to have a resistance 80 ohms greater than the second."

I'll rewrite it another way in English first to suggest it second for an algebraic translation:


"The first resistor is to equal a resistance 80 ohms more than (or in addition to) the second resistor."

R1 = R2 + 80 \(\displaystyle \ \ \ \ \ \ \ \) <------ This is where the "80" can go.



Substitute this into * from above to get:


1/9 = 1/(R2 + 80) + 1/(R2) \(\displaystyle \ \ \)or


\(\displaystyle \dfrac{1}{9} \ = \ \dfrac{1}{ \ R2 \ + \ 80 \ } \ + \ \dfrac{1}{ R2}\)


Can you solve this last equation for the variable R2?

 

[RustyLeeNorris]

I think I am wrong, but I got the common denominator as 9* R2+80 and got 9 for R2?

 

[srmichael]

I think I am wrong, but I got the common denominator as 9* R2+80 and got 9 for R2?

Not quite.

\(\displaystyle \dfrac{1}{9}=\dfrac{1}{R_{2}+80}+\dfrac{1}{R_{2}}\)

\(\displaystyle \dfrac{1}{9}=\dfrac{R_{2}}{R_{2}(R_{2}+80)}+\dfrac{R_{2}+80}{R_{2}(R_{2}+80)}\)

Combine the two fractions on the right hand side, cross multiply and then solve for \(\displaystyle R_{2}\).

Hint: You will get a quadratic equation and \(\displaystyle R_{2}\) will be an integer,