Chapter 26, Problem 26.85CP

BIO Attenuator Chains and Axons. The infinite network of resistors shown in Fig. P26.83 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. 26.83 is V_ab, then the potential difference between points c and d is V_cd = V_ab/(1 + β), where β = 2R₁(R_T + R₂)/R_TR₂ and R_T, the total resistance of the network, is given in Challenge Problem 26.83. (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 V₀, show that the potential difference between the upper and lower wires n segments from the left end is V_n = V₀/(1 + β)ⁿ. If R₁ = R_2, how many segments are needed to decrease the potential difference V_n to less than 1.0% of V₀ (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. P26.83 represents a short segment of the axon of length Δx. The resistors R₁, 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 R₂. For an axon segment of length Δx = 1.0 µm, R₁, = 6.4 × 10³ Ω and R₂ = 8.0 × 10⁸ Ω (the membrane wall is a good insulator). Calculate the total resistance R_T and β 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 m but only about 10⁻⁷ 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 mm? (c) 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 m/s. In situations where faster response is required, axons are covered with a segmented sheath of fatty myelin. The segments are about 2 mm long, separated by gaps called the nodes of Ranvier. The myelin increases the resistance of a 1.0-µm-long segment of the membrane to R₂ = 3.3 × 10¹² Ω. 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.