Question

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.

Name: Problem 93, Chapter 26: University Physics with Modern Physics
Uploaded: 2020-07-20T05:18:10-08:00
Duration: 18 min 4 s
Channel: Ren Jie Tuieng

   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.

University Physics with Modern Physics

Hugh D. Young 13th Edition

Chapter 26, Problem 93

Instant Answer

Solved by Expert Ren Jie Tuieng

07/20/2020

Step 1

We can simplify this network by considering the right-hand side of the network as a single resistor with resistance $R_T$, the total resistance of the network. This is possible because the network is infinite, and thus the right-hand side is identical to the Show more…

Show all steps

Thanks for your feedback!

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.

Play audio

Feedback

Ren Jie Tuieng and 82 other subject Physics 102 Electricity and Magnetism educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Infinite Resistor Networks

This concept refers to electrical circuits made up of an unending repetition of resistor elements. The self-similarity of the network allows one to set up a recursive relation for the equivalent resistance, meaning that removing one segment leaves a network that is electrically identical to the original. This technique is particularly useful for determining total resistance even when the circuit extends infinitely, as the solution involves solving a quadratic equation based on a fixed point condition.

Voltage Division in Resistive Circuits

Voltage division is a fundamental principle in circuits where resistors in series divide the total voltage in proportion to their resistances. When applied to cascaded resistor networks, the voltage drop across each segment can be computed, leading to a reduction or attenuation of the signal as one moves along the chain. This mechanism underlies the operation of attenuator chains and is essential for understanding how voltage levels change from one node in the network to the next.

Attenuator Chains and Exponential Voltage Attenuation

Attenuator chains consist of repeated voltage divider segments that reduce the amplitude of the voltage progressively along the chain. The cumulative effect of these segments produces an exponential decay in the voltage, with each additional segment reducing the voltage by a constant fraction. This exponential behavior is derived mathematically by recognizing the repetitive application of the voltage division rule, making it a clear example of how discrete networks can approximate continuous decay processes.

Passive Cable Model of Axons

The passive cable model is an approach used to represent the electrical properties of nerve fibers or axons by considering them as a continuous network of resistive and capacitive elements. In this analogy, the axon is modeled as an attenuator chain where the resistors represent the internal and external ionic resistances while another resistor represents the membrane resistance. This model explains how electrical signals, like voltage pulses, decay as they propagate along the length of the axon due to the inherent resistive losses.

Myelination and Its Effects on Signal Propagation

Myelination refers to the process by which axons are insulated with a fatty layer called myelin, which increases the membrane's electrical resistance significantly. In the context of the passive cable model, higher membrane resistance means a reduced leakage of current and a smaller attenuation of the voltage signal over a given distance. Consequently, myelinated axons can propagate signals faster and with less decay compared to unmyelinated fibers, which is critical for rapid neural communication.

Recommended Videos

Transcript

00:01 The main idea of this question is to show how we can model neurons with the attenuator chain, which is an infinite chain of resistors that was previously mentioned in a previous question.

00:19 So the chain looks something like this.

00:28 Right, and it goes on and on indefinitely.

00:35 Now from the question, 90dddd, we have already figured out that the total resistance r t goes to r1 plus square root of r1 square plus two r1 r2 where r1 is the ones are the resistors that are outside and r2 is the resistance in between now we have also inferred that with an infinite series that by segregating or splitting up the series right over here we say that it's identical right to our initial series right so this series if you just consider the right hand side is the same as the entire series so we can actually simplify the right hand side over here into a single resistance of rt.

02:02 Now the first part of this question, we simply want to find what is the potential difference between the point v8b, this is a and b, which is potential difference across a and b, compared to the potential difference across c and d, where c and d are over here.

02:29 C &d.

02:34 Well, what we all need to do is we need to simplify this circuit even further.

02:42 So that this equivalent resistance over here.

02:47 We can simplify them because they are in parallel with each other, so we're going to add them up, add their reciprocals up.

02:53 So this resistance over here.

02:58 Is 1 over r2 plus 1 over rt inverse.

03:13 Now to find what is the potential across c and d.

03:19 We know that the total effective resistance in this case is this expression plus 2 times of r1.

03:31 Since both of these resistors are r1.

03:38 We can take this as the effective resistance.

03:44 Then the potential, sorry, the current running through this entire circuit, b equals to v over r.

04:00 And therefore the potential of course c and d.

04:05 She's taking the current multiplied by the resistance, which in this case is 1 over r2 plus 1 over rt plus by my effective times vab.

04:23 Now of course we are not done.

04:25 We have to simplify this very ugly expression.

04:32 So this fraction over here we can combine them and then we will inverse them.

04:41 So it becomes r2rt over r2 plus rt.

04:51 This is after inversing and the effective resistance equation we also do the same simplification what we're going to do is we're going to divide this r2 rt over r2 plus rt into the denominator divided from the numerator as well as the denominator we should get vab 1 plus 2 times of r1 divided by the fraction.

05:44 So simply inverse the fraction again.

05:53 This is what you get.

05:56 And this part of the expression is simply given as beta.

06:03 In the question, this is the expression for beta.

06:06 And so this is vab over 1 plus beta.

06:18 Now to consider, subsequent segments, what are the different potentials? we can do the same process.

06:29 That is, we cut off the left -hand side and then assume the right -hand side is still equals to the total because it's an infinite series and we do the same process.

06:40 So basically if you were to have infinite segments, we start off with v -0 at the starts, imagine that there are resistors in between.

06:57 We have found that v1 is simply equals to v0 times 1 over 1 plus beta.

07:05 That's good.

07:08 Now we can ignore v0 and just focus on v1 and v2.

07:15 Right, there are still the same infinite series even if we take out v0 and we know that we can do the same process of combining the right -hand side into an r total and then doing the inverse to combine the parallel components of the circuits.

07:36 Then we do the same process again and we will be able to find v2 equals to v1 times 1 over 1 plus beta, which is equal to substituting v1 to b equals to this equation.

07:52 We get v0 over 1 plus beta square.

07:57 And we see a pattern.

07:58 So v3 must be equals to v0 over 1 plus beta cube...

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace