ladder network.attenuator chain?

by asiangrrl
Tags: chain, ladder, networkattenuator
asiangrrl is offline
Nov12-07, 01:46 AM
P: 15
1. The problem statement, all variables and given/known data

I apologize for my crappy diagramming.

The problem text exactly: "Some important kinds of networks are infinite in extent. The figure shows a chain of series and parallel resistors stretching off endlessly to the right. The line at the bottom is the resistanceless return wire for all of them. This is sometimes called an attenuator chain, or a ladder network. The problem is to find the "input resistance," that is, the equivalent resistance between terminals A and B. Our interest in this problem mainly concerns the method of solution, which takes an odd twist and which can be used in other places in physics where we have an iteration of identical devices (even an infinite chain of lenses, in optics). The point is that the input resistance which we do not yet know - call it R - will not be changed by adding a new set of resistors to the front end of the chain to make it one unit longer. But now, adding this section, we see that this new input resistance is just R_1 in series with the parallel combination of R_2 and R. We get immediately an equation that can be solved for R. Show that, if voltage V_0 is applied at the input to such a chain, the voltage at successive nodes decreases in a geometric series. What ratio is required for the resistors to make the ladder an attenuator that halves the voltage at every step? Obviously a truly infinite ladder would not be practical. Can you suggest a way to terminate it after a few sections without introducing any error in its attenuation?"

Whew, that's a mouthful, sorry you have to read through that. Basically, if I'm correct, I'm find the R_eq (R?) at each node which will follow some pattern, and apply that to find the voltage at each node. If I do the first few, they should show some geometric series pattern, which I can apply to find a ratio required to halve the voltage at every step. (I have no idea on the last question )

2. Relevant equations

R_series = R_1 + R_2 +...+R_n
R_parallel = (1/R_1 + 1/R_2 +...+1/R_n)^-1

3. The attempt at a solution

What I don't understand, which is stopping me from starting the problem, is how I'm supposed to find the R_eq. The bolded part about how R somehow affects itself (?) confuses me.

If I ignore that then:
R_eq of one "unit" = R_1 + R_2
R_eq of two "units" = R_1 + (1/(R_1 + R_2) + 1/R_2)^-1 = (2 + R_b)*R_eq of one "unit"
R_eq of three "units" = really messy

If I consider that statement:
R_eq (of one unit?) = R_1 + (1/R_2 + 1/R_eq)^-1 = eventually something quadratic that looks like (R_1 + sqrt(R_1^2 + 4R_1R_2))/2 (is my interpretation correct at all?)
R_eq of two "units" = even messier quadratic

I'd love a kick in the right direction as to what my Rs are supposed to be, I can definitely figure out the rest of the problem from there (except for the last question) I think. I don't how R "isn't changed by adding a new set of resistors" and yet "the new resistance will be ..." but I figure that both ways of trying have to be wrong since everything is coming out so messy. Thanks a bunch for reading through this.
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catkin is offline
Nov12-07, 06:56 AM
P: 218
Neat question!

Try redrawing the diagram the way the question explains the effect of adding another rung to the ladder: swivel R1 until it is vertically above R2. What have you got now? A resistance of R1 in series with (R2 and R in parallel). The first answer, R, is simply the resistance of that arrangement of resistors ...

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