BIO Attenuator Chains and Axons. The infinite network of resistors shown in Fig. $\mathrm{Pz} 6.91$ is known as an attenuator chain, since this chain of resistors causes the potential difference between the upper and lower wires to decrease, or attenuate, along the length of the chain. (a) Show that if the potential difference between the points $a$ and $b$ in Fig. 26.91 is $V_{a b},$ then the potential difference between points $c$ and $d$ is $V_{c d}=V_{a b} /(1+\beta),$ where $\beta=2 R_{1}\left(R_{\mathrm{T}}+R_{2}\right) / R_{\mathrm{T}} R_{2}$ and $R_{\mathrm{T}},$ the total resistance of the network, is given in Challenge Problem 26.91 . (See the hint given in that problem.) (b) If the potential difference between terminals $a$ and $b$ at the left end of the infinite network is $V_{0}$ , show that the potential difference between the upper and lower wires $n$ segments from the left end is $V_{n}=V_{0} /(1+\beta)^{n} .$ If $R_{1}=R_{2},$ how many segments are needed to

decrease the potential difference $V_{n}$ to less than 1.0$\%$ of $V_{0} ?(\mathrm{c})$ An infinite attenuator chain provides a model of the propagation of a voltage pulse along a nerve fiber, or axon. Each segment of the network in Fig. $\mathrm{P} 26.91$ represents a short segment of the axon of

length $\Delta x .$ The resistors $R_{1}$ represent the resistance of the fluid inside and outside the membrane wall of the axon. The resistance of the membrane to current flowing through the wall is represented by $R_{2} .$ For an axon segment of length $\Delta x=1.0 \mu \mathrm{m},$ $R_{1}=6.4 \times 10^{3} \Omega$ and $R_{2}=8.0 \times 10^{8} \Omega$ (the membrane wall

is a good insulator). Calculate the total resistance $R_{\mathrm{T}}$ and $\beta$ for an infinitely long axon. (This is a good approximation, since the length of an axon is much greater than its width; the largest axons in the human nervous system are longer than 1 $\mathrm{m}$ but only about

$10^{-7} \mathrm{m}$ in radius.) (d) By what fraction does the potential difference between the inside and outside of the axon decrease over a distance of 2.0 $\mathrm{mm}^{9}(\mathrm{e})$ The attenuation of the potential difference calculated in part (d) shows that the axon cannot simply be a passive, current-carrying electrical cable; the potential difference must periodically be reinforced along the axon's length. This reinforcement mechanism is slow, so a signal propagates along the

axon at only about 30 $\mathrm{m} / \mathrm{s}$ . In situations where faster response is

required, axons are covered with a segmented sheath of fatty myelin. The segments are about 2 $\mathrm{mm}$ long, separated by gaps called the nodes of Ranvier. The myelin increases the resistance of a $1.0-\mu \mathrm{m}$ -long segment of the membrane to $R_{2}=3.3 \times 10^{12} \Omega .$ For such a myelinated axon, by what fraction does the potential difference between the inside and outside of the axon decrease over the distance from one node of Ranvier to the next? This

smaller attenuation means the propagation speed is increased.