HW+3+solutions - 1 125:201 INTRODUCTION TO BIOMEDICAL...

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Unformatted text preview: 1 125:201 INTRODUCTION TO BIOMEDICAL ENGINEERING, Fall 2010 Bioelectrical Phenomena and Bioinstrumentation Instructor: Nada Boustany HOMEWORK #3- Due Thursday Dec. 9 SOLUTIONS REMINDER: EXAM #3 is on MONDAY DEC. 13, 2009, 8:40am-10am. RC circuit model of the membrane. Problem 1: In class on Monday Nov. 30, we wrote an expression for the voltage across the membrane at rest. In this problem you are asked to derive this expression by showing all the details leading to the answer. a- Please draw a circuit model of the cell membrane at rest (please ignore the currents through the Na + /K + ATP pumps). V m out in R Na C +Q V Na i Na i C i R K R Cl V K V Cl i K+, pump i Na+, pump Node A ATP driven sodium potassium pumps i K i Cl i K+, pump i Na+, pump b- Please show (i.e. please show all the details leading to the answer) that the voltage across the membrane is − + + − − + + + + + + + + = Cl K Na Cl Cl K K Na Na rest R R R R V R V R V V 1 1 1 At node A in this circuit, Kirchoff’s current law gives: i K + i Cl + i Na- i Na, pump + i K,pump- i C- i= 0 2 Since the membrane is at rest, i=0. Also dV/dt=0 Æ i C =0 We also simplified the problem by ignoring the effects of Na + /K + ATP pumps. This means that i Na, pump << i Na and that i K,pump <<i K. Thus, Kirchoff’s current law at node A now gives: i K + i Cl- i Na = 0 (Equation #1) Applying Kirchoff’s voltage law at three different loops gives the following addition equations: At loop1: +V m- V Na + i Na R Na = 0 (Equation #2) At loop2: -V Na + i Na R Na + i K R K + V K = 0 (Equation #3) At loop3: -V Na + i Na R Na + i Cl R Cl + V Cl = 0 (Equation #4) V m out in R Na C +Q V Na i C i R K R Cl V K V Cl Loop 3 i Na i Cl V m out in R Na C +Q V Na i C i R K R Cl V K V Cl Loop 2 i Na i K V m out in R Na C +Q V Na i Na i C i R K R Cl V K V Cl Loop 1 3 To get these equations we used the following conventions: Solving the four equations for V m , i NA , i K , and i Cl : Equ. 3 gives: k K Na Na Na K R V R i V i − − = Equ. 4 gives: Cl Cl Na Na Na Cl R V R i V i − − = Substituting for i K and i Cl into Equ. 1 : K Na Cl Na Cl K K Cl K Na Cl K Cl Na Na K Cl K Na Cl K Cl Na K Na Na Cl Na Na Cl K Na Cl K K Cl K Na Na K Na Cl K Cl Na Na Cl Na Na R R R R R R R V R V R V R V i R V R V R V R V R R i R R i R R i R R R V R R i R V R V R R i R V i + + − + − = → − + − = + + → − − + − − = Substituting for i Na into Equ. 2 : K Na Cl Na Cl K Na K Cl Na Cl K Cl K Na K Na Cl Na Cl K Na K Cl Na K Na Na Cl K Na Cl Na K Na Na Cl Na Na Cl K Na Na K Na Cl Na Cl K K Cl K Na Cl K Cl Na Na Na Na Na m R R R R R R R R V R R V R R V R R R R R R R R V R R V R R V R R V R R V R R V R R V R R R R R R R R V R V R V R V V R i V V + + + + = + + + − + − + + = ⋅ + + − + − − = − = Rearranging and factoring top and bottom by R K R Na R Cl , we get: +- +V= V A-V B A B R +- +V= V A-V B A B i R +V R = V A-V B A B C i C +V C = V A-V B +Q=CV C A B +Q Circuit conventions used in class + 4 − + + − − + +...
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This note was uploaded on 10/11/2011 for the course BIOMEDICAL 201 taught by Professor Berth during the Fall '10 term at Rutgers.

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HW+3+solutions - 1 125:201 INTRODUCTION TO BIOMEDICAL...

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