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I have stumbled upon a given question I really have a hard time to solve. Basically I need to find an equivalent resistance in some form of "ladder" configuration. Where the chain is an infinite sequence of resistors.

ladder

I have really no good idea how to find this equivalent resistance. Trying the old fashioned rule of parallel and resistors in series I came to a very messy formula: $$R + \frac{1}{\frac{1}{R} + \frac{1}{R+\frac{1}{\frac{1}{R}+....}}}$$

I know however that the solution should be much more simple.

Now I tried using Kirchhoff's loop rule. Which states that the power difference in a closed loop must be 0. Naming the "potential currents" between AB $I_1$, between BC (through the single resistance) $I_2$ and the current "from B to the right" $I_3$ Considering the loop containing BC & the rest of the structure this rewrites to: $$I_1 = I_2 + I_3$$ $$I_3 \cdot R - I_2 \cdot R_{eq} = 0$$

The problem is, 2 variables, 2 functions doesn't really bring me closer to an answer :(. What am I missing?

MrAP
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paul23
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3 Answers3

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HINT:

Notice that $$R_{eq}=R+\frac1{\frac1R+\frac1{R_{eq}}}$$

Ruslan
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    From this indeed follow the answer quickly. Now to understand why you are allowed to make such a statement in this case. I guess it comes from the fact the sequence is infinite in size; as such "rewriting" the first area ABCD does not affect $R_{eq}$ (In other words $R_eq$ is the same even after removing an occurence). Is this the correct assumption to make in such a question? – paul23 Aug 09 '13 at 23:23
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    Yes, it is correct provided the continued fraction converges ($R_{eq}$ is finite). And it does converge because it's shunted by first $R$. – Ruslan Aug 09 '13 at 23:25
  • +1: Very cool argument, and to think I was sitting here trying to solve recursion relations :) – joshphysics Aug 09 '13 at 23:35
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You can divide the circuit in n stages, the first stage on the right will show resistance r_0=2R and then you will have to move to the left. At the $n$th stage you will have unknown resistance $r_n$ and at the $(n+1)$th stage you will have resistance: $$r_{n+1}=R+\dfrac{Rr_n}{R+r_n},$$ being the series of $R$ with the parallel between $R$ and $r_n$. This is a recursive succession which, if converges to a limit $L$, it will show $L=r_n=r_{n+1}$ when $n$ tends to infinity. Substituting you will obtain: $$L=R+\dfrac{RL}{R+L}$$ providing the second grade equation $L^2-RL-R^2=0$. The desired limit $L$ is thus the $R_{eq}$ that you are searching: $$R_{eq}=L=\dfrac{R}{2}(1+\sqrt{5})$$

fibonatic
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Antonio
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its very simple, follow these steps

1)as the circuit tends to infinity, consider the equivalent resistance to be R(e) 2)consider the parallel resistors i.e., BC and the rest to the right of BC. these are in series with AB 3)now the total resistance to the right of BC will be equal to the effective resistance of the total circuit R(e), as the circuit tends to infinity _note: this is possible only in infinite ladder series. 4)the effective resistance will be R(e)= r + {R(e)*r}/{R(e)+r} 5) on solving this you will get a quadratic equation which has a +ve and a -ve root. The +ve root gives the answer. note: here {R(e)*r}/{R(e)+r} gives the effective resistance of the resistors in parallel as mentioned above.

B.Ashwin
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