Effective Resistance of an Infinitely Long Ladder of Resistors

[malindenmoyer]

Find the effective resistance (resistance between a and b) of an infinitely long ladder of resistors, as shown in the figure, each having resistance R.

http://people.tamu.edu/~malindenmoyer/fig.jpg

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 R₁ in series with the parallel combination of R₂ and R. We get immediately an equation that can be solved for R.

 

[rl.bhat]

If you consider Req. at a'b' line, then R and Req are in parallel and other 2 R are in series with them. Simplify the circuit and solve the quadratic to find Req..

 

[malindenmoyer]

still confused, perhaps a more detailed hint?

 

[collinsmark]

Hello malindenmoyer,

A very important part of this is your quote,

"[R_eq] will not be changed by adding a new set of resistors to the front end of the chain to make it one unit longer."

So take R_eq as the equivalent resistance looking to the right of line a'b'.

Model R_eq as a single resistor if you want, at this point. Now find the equivalent resistance once you add in the extra three R resistors, between points ab and a'b'. Find the equivalent resistance of that (you'll have 4 resistors total. 2 series R, in series with an R and an R_eq in parallel).

But you know from your above quote, that the equivalent resistance of that new circuit (above paragraph) won't change the equivalent resistance before the 3 new resistors were added. So you can set the whole thing equal to R_eq. Then solve for R_eq.

 

[malindenmoyer]

Okay, from both of the responses, I was able to come up with this expression, am I on the right track?

$$R_{eq}^2-\frac{2R^2-1}{R}R_{eq}-1=0$$

If correct, do I use the quadratic formula to solve?

 

[collinsmark]

Okay, from both of the responses, I was able to come up with this expression, am I on the right track?

$$R_{eq}^2-\frac{2R^2-1}{R}R_{eq}-1=0$$

If correct, do I use the quadratic formula to solve?

Yes, you will be using the quadratic equation/formula for this one. But I don't think your equation is correct.

Hint: Each term in the equation (in the form aR_eq² + bR_eq +c = 0) needs to have units of $$\Omega ^2$$ (meaning 'a' is unitless, 'b' has units of $$\Omega$$, and 'c' has units of $$\Omega^2$$).

 

[rl.bhat]

Req = 2R + [ R*Req/(R + Req).]
Now simplify and solve for Req.