# HELP! Resistors Parallel/series

1. Mar 5, 2005

### 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?

2. Mar 5, 2005

### faust9

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

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.

3. Mar 5, 2005

### 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.

4. Mar 5, 2005

### 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.