Equivalent resistance between two points

[Vibhor]

Homework Statement

Homework Equations

The Attempt at a Solution

[/B]
All the resistors are connected in parallel across A and B .The equivalent resistance is 1/R_eq= (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

 

[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.

 

[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. T_n=?

 

[haruspex]

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.

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.

 

[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.

 

[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.

 

[cnh1995]

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.

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.

 

[rude man]

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.

That sounds like the way to go all right.

 

[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.

 

[Hamid438]

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

 

[cnh1995]

Welcome to PF!

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

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