Calcuating the resistance of a circuit.

[neener]

(Solved) Calcuating the resistance of a circuit.

Hey everyone, long time lurker, first time poster. Glad to be here!

I have a few physics problems that I need assistance with. I may create another thread later. For now, here's this:

Homework Statement

Consider the following circuit:

[PLAIN]http://img192.imageshack.us/img192/2285/phystopic1.gif

(1) Redraw the circuit

(2) Using the circuit in (1), what is the net resistance between points A and B? Each resistor has the same resistance R. (Show your derivations)

(3) What is the resistance between B and C? Explain.

(4) A [tex]\epsilon[/tex] Volts emf is connected to terminals A and B. Derive an expression for the current in one of the resistors.

Homework Equations

Series: R_total=R1 + R2 + ... + Rn
Parallel: 1/R_total = 1/R1 + 1/R2 + ... + 1/Rn

The Attempt at a Solution

So this is where I show you that I'm actually attempting to do some work rather than just trying to get some handouts. I've redrawn a couple (much simpler) circuits before, but this one just baffles me! Can anybody give me any tips or pointers on how to even start this so I can start attempting the rest of the problem? I'd really appreciate any help. Thanks!

 

[praharmitra]

Can you explain what a parallel connection of two resistors is?

 

[gneill]

What's the resistance between any two adjacent vertices of the octagon?

 

[neener]

Can you explain what a parallel connection of two resistors is?

Each end of the two resistors shares the same connection, causing each to have the same potential difference, right?

What's the resistance between any two adjacent vertices of the octagon?

Isn't it just zero?

 

[praharmitra]

Each end of the two resistors shares the same connection, causing each to have the same potential difference, right?

Thats right! Now can you see that the two ends of all your resistors indeed have the same connection? They are all in parallel!

 

[gneill]

GN - "What's the resistance between any two adjacent vertices of the octagon?"

Neener - "Isn't it just zero?"

You need to be sure. What are your arguments for and against if you're not sure?

 

[neener]

Thats right! Now can you see that the two ends of all your resistors indeed have the same connection? They are all in parallel!

Egads! I see that. However, I don't think I understand what implications that would have when, for example, measuring the resistance between points A and B, and B and C.

Does the redrawn circuit just look like this?

[PLAIN]http://img844.imageshack.us/img844/4661/phystopic2.gif

If this is the case, then where are points A, B, and C now? I'm guessing A is just one of the main connectors and B and C are one of the connectors on anyone of the resistors?

I guess I'm a little thrown off because I'm so used to there being a source (e.g. a battery) and redrawing it from there. I suppose in this case, it doesn't really matter (Lots of symmetry going on here, too).

 

[gneill]

Where A,B, and C live in your new diagram.

 

[praharmitra]

If ever in doubt, always draw a battery with potential V to the two points b/w which resistance is to be calculated. Find the current drawn from the battery. The resistance is V / I. Fullproof (albeit lengthy) way to solve it all. :)

 

[neener]

You need to be sure. What are your arguments for and against if you're not sure?

Good point. I say zero because for any two given points on that octagon, there always exists a path that need not go through any resistors.

Edit: I just saw your addition to my diagram (thank you!) and my guess appears to be correct.

 

[neener]

Alrighty.

So part (1) of the question is complete.

(2): Looking at the re-drawn circuit, the net resistance between A and B (given that each resistor has the same resistance R) should be:

1 / (8 x (1/R)) which is equal to R / 8

(3): There is no resistance between B and C. That path bypasses all the resistors.

(4): We want to find the current of one of the resistors when given e Volts between A and B. We use Ohm's law (I = V/R). Since all the resistors are sharing the current, we just get:

I_resistor = e/R

Does everything look good?

 

[gneill]

Looks good.

 

[neener]

Thanks a bunch for the help, gneill and praharmitra!