Question

BIO Attenuator Chains and Axons. The infinite network of resistors shown in Fig. $\mathrm{P} 26.83$ 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 $V_{a b},$ 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 26.83. (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 $V_{0}$, 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 $V_{0}$ ? (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 $\Delta x$. 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 $2.0 \mathrm{~mm} ?$ (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.

Name: Problem 85, Chapter 26: University Physics with Modern Physics In SI Units
Uploaded: 2023-01-25T14:49:20-08:00
Channel: Lainey Roebuck

   BIO Attenuator Chains and Axons. The infinite network of resistors shown in Fig. $\mathrm{P} 26.83$ 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 $V_{a b},$ 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 26.83. (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 $V_{0}$, 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 $V_{0}$ ?
(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 $\Delta x$. 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 $2.0 \mathrm{~mm} ?$
(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.

University Physics with Modern Physics In SI Units

Hugh D Young; Roger… 15th Edition

Chapter 26, Problem 85

Instant Answer

Solved by Expert Lainey Roebuck

01/25/2023

Step 1

We know that $R_{eq} = R_2R_T / (R_2 + R_T)$, so we can substitute this into the equation to get $V_{cd} = V_{ab} / (1 + \beta)$, where $\beta = 2R_1(R_T + R_2) / R_TR_2$. Show more…

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BIO Attenuator Chains and Axons. The infinite network of resistors shown in Fig. $\mathrm{P} 26.83$ 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 $V_{a b},$ 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 26.83. (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 $V_{0}$, 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 $V_{0}$ ? (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 $\Delta x$. 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 $2.0 \mathrm{~mm} ?$ (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.

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Key Concepts

Equivalent Resistance in Complex Networks

In these types of problems, calculating the equivalent resistance of an infinitely long network is essential. This involves setting up and solving equations where the effective resistance of an appended segment is equal to the combined resistance seen by the source, which is a fundamental concept in circuit analysis and critical for determining how voltage divides across the network.

Myelination and Its Effect on Signal Propagation

Myelination refers to the insulation provided by myelin sheaths covering some axons, which significantly increases the membrane resistance. This increased resistance reduces the voltage attenuation between the nodes (gaps in the myelin coating) and thereby enhances the speed at which a nerve signal propagates. The concept illustrates how biological adaptations can optimize electrical signal conduction over long distances.

Axon as an Electrical Cable Model

Modeling an axon as a passive electrical cable uses the resistor network analogy to represent the flow of ionic currents along a nerve fiber. In this model, the resistors symbolize the resistance offered by the internal and external fluids and the membrane’s resistance to current flow, providing insight into how a voltage pulse decays as it travels along the axon.

Infinite Resistor Network Analysis

Infinite resistor network analysis often exploits the self-similar, repeating structure of the circuit. By considering that the total resistance of the network does not change when additional segments are added, one can form recursive equations that allow for the determination of quantities such as the total network resistance and the voltage drop across each segment.

Voltage Attenuation and Division

Voltage attenuation in resistor networks involves the use of voltage division principles, where the potential difference is partitioned across series and parallel resistive elements. In an infinite chain, a recursive method is applied to determine the effective resistance and the corresponding attenuation factor, leading to formulas that exhibit an exponential reduction in voltage over discrete segments.

Attenuator Chain

An attenuator chain is an infinite network of resistor segments designed to reduce or 'attenuate' the voltage as it propagates along the chain. The concept relies on the repeated application of voltage division across identical segments, which results in an exponential decrease of the input voltage along the length of the network.

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