Cable TV version of infinite ladder network

Question

This is a problem in a college physics textbook, and its bugging me that I can't get it.

The figure shows a circuit model for the transmission of an electrical signal, such as cable TV, to a large number of subscribers. Each subscriber connects a load resistance RL between the transmission line and the ground. The ground is assumed to be at zero potential and able to carry any current between any ground connections with negligible resistance. The resistance of the transmission line itself between the connection points of different subscribers is modeled as the constant resistance RT.

Prove the the equivilent resistance as seen by the cable tv company is the equation at the top of the image.

I found this suggestion: Because the number of subscribers is large, the equivalent resistance would not change noticably if the first subscriber cancelled his service. Consequently, the equivalent resistance of the section of the circuit to the right of the first load resistor is nearly equal to Req.

Answer 1

I'll bite.

Imagine we know the resistance of the system: $R_{eq}$. Your hint was right, if we add another section we won't change the result (because for a large number of sections, one more isn't going to change the result-think infinite sums that converge). To add another section we need to add $R_T$ in series then $R_L$ in parallel:

$R_{eq}= R_T+ \frac{R_LR_{eq}}{R_L+R_{eq}}$

Re-arranging this equation we obtain a quadratic:

$R_{eq}^2-R_TR_{eq}-R_TR_L=0$

Solving with the quadratic formula yields:

$R_{eq}=\frac{1}{2}(R_T\pm(R_T^2+4R_TR_L)^{\frac{1}{2}})$

We can discount the negative solution because a negative resistance doesn't make sense. This gives the final answer:

$R_{eq}=\frac{1}{2}[(4R_TR_L+R_T^2)^{\frac{1}{2}}+R_T]$

Which is in the form given in your question.

I have purposely skipped algebra amongest my explanation so you can have fill in whats missing instead of me typing a wall of steps - I believe these are the main ones.