What Are the Possible Values for R in a Parallel/Series Resistor Setup?

[NotaPhysicsMan]

Ok the question is:

A resistor (resistance=R) is connected first in parallel and then in series with a 2.00 ohm resistor. A battery delivers five times as much current to the parallel combination as it does the series combination. Determine the two possible values for R.

So let's see: I can use I=V/R

For series it's I=V/(R+2)

For parallel it's 5I=V/(1/R+1/2ohm) Ok, now I equate the Currents (I)

and I get V/(R+2)=V/5(1/R+1/2). I can cancel out the V's and I'll have:

Now I solve for R?

Am I on the right track?

 

[faust9]

Your parallel equation is wrong:

R_{eq}=(\frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n})^{-1}

So if you say I=\frac{V}{R_{eq}} you end up with:

I=\frac{V}{(\frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n})^{-1}}

or more simply:

I={V}{(\frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n})}

Your thought process is correct though.

Good luck.

 

[so-crates]

Your equation for equivalent resistnace in a parallel circuit is wrong. The correct equation is:

\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}

Otherwise you seem to be on the right track.

 

[NotaPhysicsMan]

Ok this is weird. Ok so jumping to the equating of the two equation step...

I get :

V/(R+2)=V/5 x (1/R+1/2). Now, I cancel the Vs and try to solve for R

1/(R+2)=1/5 x (1/R+1/2). Now I bring the 5 over on the otherside and now I will solve and get a quadratic equation:

R^2-6R+4=0. Solving for R's I get R=5.236 and R=.7639 ohm.

It looks right? Thanks.