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Chap 4 Lecture 6

The document discusses the circuit analysis of cell membranes using Ohm's and Kirchhoff's laws, focusing on the role of potassium (K+), sodium (Na+), and calcium (Ca2+) in resting and action potentials. It presents qualitative and quantitative analyses of a circuit involving two resistors and batteries, demonstrating how the potential difference (PD) across the network is influenced by the resistances and battery values. The findings emphasize that the battery with the lowest series resistance is the most expressed in the network, regardless of the absolute values of the components.

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5 views3 pages

Chap 4 Lecture 6

The document discusses the circuit analysis of cell membranes using Ohm's and Kirchhoff's laws, focusing on the role of potassium (K+), sodium (Na+), and calcium (Ca2+) in resting and action potentials. It presents qualitative and quantitative analyses of a circuit involving two resistors and batteries, demonstrating how the potential difference (PD) across the network is influenced by the resistances and battery values. The findings emphasize that the battery with the lowest series resistance is the most expressed in the network, regardless of the absolute values of the components.

Uploaded by

heer41162
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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CIRCUIT ANALYSIS FOR CELL

MEMBRANE
Book: ELECTROGENESIS OF BIOPOTENTIALS IN THE CARDIOVASCULAR SYSTEM
by Nicholas Sperelakis

Using Ohm's and Kirchhoffs laws and logic, It is possible to see why in the nerve or muscle cell
the membrane K+ dominates the resting potential, and Na+ or Ca2+, or both, dominates the peak
of the action potential.

Before analyzing the circuit rigorously, we can consider three conditions and make some
qualitative judgments:

1) If the left resistor (R1) equals the right resistor (R2) in Figure VI-AC-l (regardless of their
absolute values), the PD across the network is +150 V (upper terminal positive with respect to
the lower terminal), i.e., halfway between both batteries, since both should be equally expressed.
2) If R2 is made infInite (e.g., open circuit in branch 2) and Rl is finite, then the PD is exactly
+ 100 V, since the left battery in Figure VI-AC-l (E1) cannot be expressed at all.

3) If Rl is much less than R2 , then the PD approaches + 100 V, since E1 is dominant.

The circuit can also be analyzed quantitatively. For example, if R1 = 10 0 and R2 = 9900, the
exact PD may be reasoned from the following analysis. The current (I) has one magnitude, i.e., is
constant, throughout this simple closed circuit, but the current flows upward in branch 2 and
downward in branch 1. This occurs because the right battery in Figure VI-AC-l (E2) is larger
than El , and so the net driving force for the net current is in the direction as indicated in Figure
VI-AC-l.

Therefore the voltage drops produced across Rl and R2 are in opposite polarities, as shown in
Figure VI-AC- 1. The voltage drop across R1 adds to E1 to make a greater PD across branch 1,
like two batteries in series (+ - + -). In contrast, the voltage drop across R2 subtracts from E2 to
make a smaller PD, like two batteries back to back (- + + -). Therefore the following two
equations can be written for the PD across branch 1 [(PD)1] and across branch 2 [(PD)2].

= − .. −1

= − .. −2

To solve these equations we must first calculate the current (I). The net driving force for I is
equal to E2-E1; hence from Ohm's law the net current is equal to (E2 - E1) divided by the total
resistance (R1 + R2)

−
= … −3
+

200 − 100
=
990 + 10
100
= = 0.1
1000
Replace this value in equation AC-1 and AC-2. For PD1 current is flowing in opposite direction.
So, sign will be negative.

= −

= 100 − −0.1 10

= 100 + 1 = 101
Same way,

= −

= 200 − 0.1 990

= 200 − 99 = 101

Because the two branches are connected by zero resistances, i.e., they are effectively the same
points, (PD)1 must equal (PD)2.

= … −4

To summarize, it is quantitatively demonstrated that the battery with the lowest series resistance
is the battery most expressed across this network. This analysis holds true regardless of the
absolute values of the resistances or batteries or the polarity of each battery.

Other methods of circuit analysis can be used to calculate the PD across such a network, but this
method is one of the simplest. The following equation, analogous to the chord conductance
equation, can also be derived and used to calculate the PD across the network

= +
+ +

10 990
= 200 + 100
10 + 990 10 + 100
2000 99000
= +
1000 1000
= 2 + 99 = 101

This equation again emphasizes the point that it is the relative resistance that determines which
battery is most expressed.

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