Equivalent resistance between two points

1. Jun 2, 2017

Vibhor

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

2. Relevant equations

3. The attempt at a solution

All the resistors are connected in parallel across A and B .The equivalent resistance is 1/Req= (1/R) +(1/2R) + (1/6R) + (1/12R) + ....

I tried to work with few terms and it seems the effective resistance tends to zero as terms increase . But C) is not the correct option .

Now I am not sure how to calculate this sum . Am I missing something very simple ?

Thanks

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2. Jun 2, 2017

magoo

Calculate the result for up to 4 resistors in parallel.

Look at the results for 1, 2, 3 and then 4 resistors in parallel.

For 1 resistor, Req = R.

For 2 resistors, Req = 2/3 R

continue for the next two.

Observe how much of a change you see from 1 to 2, 2 to 3, etc.

It should be decreasing but it doesn't go to 0.

Another observation is when the added resistor in parallel gets to be about 1/20 R, it doesn't add much to Req.

3. Jun 3, 2017

cnh1995

Zero is not the answer for sure, as you can see the values of the parallel resistances go on increasing.

Leave the first R alone. What can you say about the sequence 2R, 6R, 12R, 20R...?
What is the expression for this sequence?
T1=2R, T2=6R.....and so on. What is the general expression for this sequence i.e. Tn=?

4. Jun 3, 2017

haruspex

That all may be suggestive but proves nothing. The first step is to figure out the general form of the sequence 1, 2, 6, 12, 20....
Actually, it doesn't have a guessable one unless you omit the initial 1. That seems to be a mistake by the question setter.

5. Jun 3, 2017

magoo

My suggestion was merely to calculate the results for up to 4 resistors in parallel. If you look at the values or plot the Req versus # of resistors, you'll see that the curve decreases to a value other than 0. You can do it for the 5 values shown also.

If you haven't figured out the series, the fact that the 20R and greater terms are insignificant in the total makes this a reasonably acceptable approach. If the higher terms had greater impact on Req, then this would not be acceptable.

6. Jun 3, 2017

rude man

I agree with haruspex, the sequence is far from obvious.

There is a website called "Car Talk" run nominally by two MIT graduates (one sadly deceased). This is the sort of puzzle they post. The answer is usually irrelevant to mathemaics to any perceptible extent. This one looks like another such.

7. Jun 3, 2017

cnh1995

I think this approach is simpler (and smarter!). Simply calculate the equivalent resistance up to 4 or 5 resistors and compare that value with the options. The closest option would be the correct one.

Last edited: Jun 3, 2017
8. Jun 3, 2017

rude man

That sounds like the way to go all right.

9. Jun 3, 2017

epenguin

You can only give a unique answer if you assume going beyond what we are told, that this series which we don't know is convergent. If it is not, if it is divergent, equivalent resistance is 0 (c) .
Assuming then that the series is convergent we can answer the question since we don't need to know what the equivalent resistance is, only what it isn't. It's fairly obviously isn't b. By working out the equivalent resistance of the part we can see we can decide between the two remaining possibilities.

10. Jun 5, 2017

Hamid438

Sum of 1/r approaches 2R
Rt=R/2 approximate.

11. Jun 5, 2017