# Electric Circuit Problem

1. Oct 12, 2011

### aguycalledwil

1. The problem statement, all variables and given/known data

A distant telephone, whose resistance is 300 ohms, is connected to the exchange 10km away by a pair of wires whose resistance singly is 50 ohms/km. A twig falls across the wires at a certain point producing the effect of a resistance, R, between the wires at that point. Measurements made at the exchange show a resistance between the ends of the wires of 130 ohms which rises to 160 ohms when the distant telephone is disconnected.

How far from the exchange is the twig to be found and what is the value of R?

2. Relevant equations

As far as I can tell, all that is needed is..

SERIES
∑R = R1 + R2 + R3 etc..

PARALLEL
1/∑R = 1/R1 + 1/R2 + 1/R3 etc..

3. The attempt at a solution

I tried to get two simultaneous equations in R and x (with x being the distance of the twig to the exchange)

So for my first equation, I said.. ∑R=R+(5x*2)
so.. 160=R+10x

For my second equation, I said.. 1/∑R = 1/(R+10x) + 1/(300+2*5*(10-x))
so.. 1/130 = 1/(R+10x) + 1/(400-10x)

However solving these simultaneous equations is proving brutal - I can't get the second one in a nice form. Any help?

Thanks!

2. Oct 12, 2011

### Staff: Mentor

Are you sure that all the numbers given in the problem statement are correct? I don't find realistic solution values for distance and R (one or the other turns out to be negative).

3. Oct 13, 2011

### aguycalledwil

Yeah, I'm 100%.

I also got a negative answer at one point but assumed it to be wrong. This is an extension question so it's meant to be hard but i'm starting to wonder if it's possible..

4. Oct 13, 2011

### Staff: Mentor

We are told the line as far as the twig, plus the twig's resistance, amount to 160 ohms. So at 100 ohms per km, this indicates the distance is no farther than 1.6 km. The twig's resistance must be less than 160 ohms. So the extra that is placed in parallel with the twig when the phone is in place, must be at least 300 + 8.4*100 ohms, i.e., 8700 ohms. We are lead to believe that 8700 ohms in parallel with the resistance of the twig, is significant.

8700 ohms paralleling the twig of between 0 and 160 ohms is not significant; it can't bring the overall resistance down to 130 ohms.

So there is no solution. (Maybe a negative resistance.)

Last edited: Oct 13, 2011
5. Oct 13, 2011

### aguycalledwil

Well, I took my half solution to class today, and it turned out I was fairly close. There is in fact an answer, it is 4 km and 120 ohms. The formula I should have used for parallel resistors is ∑R = Sum of resistors/Product of resistors.

Using this, the maths becomes a lot easier..

Thanks for your help anyway!

6. Oct 13, 2011

### Staff: Mentor

Can you show the work? I'd like to see how this solves the problem.

4km at 50 Ohms per km, and two wires of that length, yields a minimum loop resistance of 400 Ohms. That exceeds the required 130 Ohm value.

7. Oct 13, 2011

### Staff: Mentor

:surprised: :surprised: :surprised: Err, no. It isn't.
:surprised: I guess we will now have to rewrite all the textbooks!
I can't argue with that logic!
Sorry we had it so wrong. :yuck:

Lesson number one: you must do it the way the teacher wants.

8. Oct 14, 2011

### Staff: Mentor

If you have two resistors, each of 1 ohm, and you connect them in parallel, what is the equivalent resistance--according to your easy-to-use formula?

What would be the circuit resistance if you connected the two resistors in series?

9. Oct 14, 2011

### Staff: Mentor

You stated 50 ohms/km for the cable wires, yet I can see in your working you consistently used a value of 5 ohms/km. So the data was changed from the original specs?

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