J. Newman, Physics of the Life Sciences, DOI: 10.1007/978-0-387-77259-2_16, © Springer Science+Business Media, LLC 2008
Thus far in our study of electricity, we have essentially confined our attention to electrostatics, or the study of stationary charges. Here and in the next three chapters we show some of the new phenomena that arise when charges move. We begin this chapter by generalizing our discussion to allow the flow of electric charges, known as an electric current, and we give a semiempirical derivation of Ohm’s law. Electrical measurement methods and devices are described as an application of Ohm’s law. More realistic models for a capacitor are then developed in a continued study of cell membranes in which electric charge can passively leak across the mem-brane. We give an overview of nerve structure and functioning and the spatial and temporal properties of the neuron membrane potential are detailed for both the qui-escent and active states. The chapter concludes with a discussion of the electrical properties of individual ion channels as the underlying basis for membrane currents.
1. ELECTRIC CURRENT AND RESISTANCE
Although we have introduced the topic of membrane channels in the last chapter, we have not discussed the consequences of channels on the electrical properties of mem-branes. Membranes act as capacitors, storing charge and electric potential energy, but because of “leakage” of charge through channels, membranes are not the ideal capac-itors treated in the last chapter. In order to discuss more realistic models for mem-brane electrical properties we first need to introduce some concepts related to the flow of electric charge.
Figure 16.1 shows a conducting wire attached at time zero between the plates of a previously charged air-spaced capacitor. Before the wire is connected we have already seen that there is an electric field between the plates of the capacitor, but no charge flows because the air is a good electrical insulator. As soon as the wire is con-nected, there will be an electric field in the wire that will drive the free electrons toward the positive capacitor plate, discharging the capacitor. The electric current in the wire is defined as the time rate of flow of charge along the wire
(16.1)
where the direction of the current is chosen by convention as opposite to the flow of the electrons. Thus, the electric current flows from the positive to negative plates of the capacitor in our example. The SI unit for electric current is the ampere (A), given by Equation (16.1) as 1 C/s⫽ 1 A.
In our example, all of the net charge will travel through the wire very rapidly, resulting in a final uncharged capacitor. Clearly the electric current flowing in the wire is not constant in this situation because as the charge drains off the capacitor
I⫽ ¢Q
¢t ,
16
Electric Current and Cell
plates, the electric field that drives the electric charges decreases. If the initial charge on each capacitor plate was 1C and the flow of charge is complete within 1 s, then the average electric current flowing is given by Equation (16.1) as I⫽ 1 C/ 1s ⫽ 1 C/s ⫽ 1 A. But clearly the current is not constant over this 1 s, decreasing continuously as the charge is drained from the capacitor plates. We show below how to find the actual time dependent current flowing in this simple electric circuit.
Unlike the electric fields of previous chapters, the electric field driving the charges through the wire is not an electrostatic field. In fact, as we have seen, trostatic fields cannot exist within a conductor. The electric field that drives the elec-tric current, on the other hand, does exist within the conductor and is responsible for pushing the charge making up the current. This example illustrates that without a source of energy to maintain net charge on the plates of the capacitor, both the elec-tric field in the wire and the current flow rapidly decrease to zero.
After charging the capacitor in Figure 16.1, we can think of the discharging of the capacitor as the conversion of electric potential energy to the kinetic energy of the elec-trons in the wire connecting the plates. As we show at the end of this section, the kinetic energy of the free electrons making up the current is then converted into heat via colli-sions with the metal atoms of the wire. The discharging of the capacitor occurs rapidly and therefore there is only a pulse of electric current in this case. In order to maintain a flow of electric charge, an external source of energy per unit charge, traditionally called an emf (pronounced “ee em eff,” and short for the misnomer—electromagnetic force— because it is not really a force), is needed in the form of a battery or power supply.
The simple electric circuit shown in Figure 16.2 (left) consists of a battery with a uniform wire connected between its terminals. If the battery were simply a capacitor as in Figure 16.1, the initially separated positive and negative charges would quickly can-cel each other out as charge flows along the wire and there would be no further change. Batteries convert chemical energy to electrical energy to continually maintain a separa-tion of charge and supply a fixed voltage between their terminals. This is shown on the right side of the figure where the varying voltage is shown as it might be measured around the circuit, with the battery increasing the voltage each time around. A very good analogy is the flow of water due to gravity where the potential energy decreases as the water flows down hill and can only be restored by a pump of some kind, playing the role of the battery, to increase the height and thus the potential energy of the water. In our case the uniform wire of length L has a constant potential V between its ends, resulting, as we show, in a constant current flow along the wire. The constant flow of current is produced by a uniform electric field in the wire maintained by the battery and given by E⫽ V/L. Electric field lines begin on the positive (⫹) terminal and end on the negative (⫺) terminal of the battery as long as the wire has no sharp bends and is smooth.
We can understand the origin of the constant current in this case by considering a microscopic picture of a collection of free electrons in the conducting wire and the forces acting on them. In the absence of an external electric field, the thermal energy of the free electrons causes them to diffuse about in a random walk traveling at very high speeds of about 106m/s and making random collisions with the atoms of the
metal wire (see the discussion in Chapter 2). The average velocity of the electrons, as opposed to their high speed, is zero in this case and there is no net flow of charge, therefore no electric current. When an electric field is applied, superimposed on its high-speed random walk motion, a free electron will experience an acceleration (in the direction opposite to the electric field because of the negative electric charge) given by
(16.2)
where e and m are the charge and mass of the electron. This acceleration lasts until the electron makes a collision with a metal atom causing it to veer off in another random direction at high speed, accelerating again according to Equation (16.2).
a⫽ F m⫽ eE m, E I +Q –Q
FIGURE 16.1Two charged conducting plates connected by a conducting wire at time zero.
+
-voltage
Distance around circuit from battery
wire battery
wire battery
FIGURE 16.2(left) A battery with its terminals connected by a uniform wire. (right) Voltage as a function of distance around the circuit showing the decreasing voltage in the wires and the boost in voltage across the battery from chemical energy every time around the circuit loop.
The mean time between collisions, , is so short that the electrons only acquire a very slow drift velocity of about 10⫺3m/s given by
(16.3)
If the number of free electrons per unit volume, or number density, in the wire is n and the wire has a cross-sectional area A, then the net free charge in a short length of the wire l is ⌬Q ⫽ nAle (Figure 16.3). To find the current in the wire, we must divide ⌬Q by the time required for all of that charge to move a distance l down the wire, ⌬t ⫽ l/vdrift, to find (16.4)
Substituting from Equation (16.3), the electric current is
(16.5)
Defining the conductivityof the wire, an intrinsic property of the material, to be
we can rewrite Equation (16.5) as
(16.6)
where G is known as the conductance.
Solving for V, this can be rewritten in terms of the resistance R
(16.7)
where
The resistivity of the material is given by the inverse of the conductivity,
both intrinsic parameters. This definition is made in analogy with the equality between the resistance and the inverse of the conductance
except that both of these quantities are dependent on the size and shape of the mate-rial, so that they are extrinsic parameters, unlike the intrinsic parameters depending only on the nature of the material and not on any geometric parameters.
We conclude that the current flowing in a conducting wire is proportional to the potential difference applied between the ends of the wire. This linearity of current with applied voltage (Equation (16.7)) is known as Ohm’s law. A plot of the current through a wire as a function of the voltage across the wire is shown in curve A of Figure 16.4. The linear plot is characteristic of an ohmic (or linear) circuit element. Another equiv-alent statement of Ohm’s law is that the resistivity of the material remains a constant, independent of the applied voltage.
R⫽ 1 G, r⫽ 1 s , R⫽ 1 G⫽ 1 s L A⫽r L A. V⫽ IR, I⫽ sA LV⫽ GV, s⫽ne 2t m , I⫽ne 2t A mL V. I⫽ ¢Q ¢t ⫽ nAle
(l/ vdrift )⫽ nAevdrift. vdrift⫽ at ⫽eE m t⫽ eVt mL. l A vd
FIGURE 16.3Free charge in a wire of cross-sectional area A and length l traveling with a drift velocity vd.
I
V Slope=1/R A
B
FIGURE 16.4The I–V curve for an ohmic circuit element (A) and a semiconductor diode (B).
The SI unit for resistance is the ohm (⍀), where 1 V/A ⫽ 1 ⍀ (read as 1 ohm). Units for resistivity are then given as ⍀-m and for conductivity as (⍀-m)⫺1. The unit for conduc-tance, the reciprocal of resisconduc-tance, is the ⍀⫺1 which is also known as the siemens (S). Table 16.1 lists some values for resistivity of various materials. A wire made from a metal will have a very low resistance value. For example, a 1 m length of 1 mm diameter copper wire has a resistance of only 0.02 ⍀. Simple devices known as resistors (shown in Figure 16.5) are manufactured to have various resistance values. The symbol is used to represent a resistor in a schematic or circuit diagram such as the one shown in Figure 16.6. Connecting wires have negligible resistance, so that their length and shape are usually not important in a circuit diagram or in the actual circuit itself.
Table 16.1Resistivities of Various Materials (20°C) Material Resistivity, (⍀ ..m) Conductors Aluminum 2.8⫻ 10⫺8 Copper 1.7⫻ 10⫺8 Iron 10.⫻ 10⫺8 Mercury 96.⫻ 10⫺8 Silver 1.6⫻ 10⫺8 Tungsten 5.6⫻ 10⫺8 Ionic materials Water (distilled) ~2⫻ 105 Fresh water ~5⫻ 102 Sea water ~0.3 Cytoplasm ~0.5 Fatty tissue ~15 Semiconductors Germanium ~0.5 Silicon ~2.⫻ 103 Insulators Air (dry) 4⫻ 1013 Glass 1010⫺ 1014 Rubber 1013⫺ 1016
Example 16.1 How much electric current flows through water contained in an insulating tube 10 cm long and 5 cm in diameter when a 100 V potential differ-ence is applied across the ends of the tube using electrodes inserted at either end? Ignore any complications from the metal electrode–water contact and do the calculation using the three entries in Table 16.1 for different purities of water.
Solution:The current that will flow is given from Ohm’s law by I⫽ V/R, where R is the resistance between the two electrodes supplying the 100 V potential difference. Using the relation between resistivity and resistance, and the dimen-sions of the water tube, we find that
FIGURE 16.5An assortment of resistors.
R1
R2
FIGURE 16.6A simple circuit diagram showing a battery connected to two resistors, one wired after the other.
Ohm’s law is not a fundamental law on par, for example, with Newton’s laws. It is a heuristically derived statement that the current and voltage are proportional in a conductor. Many electrical components, such as diodes, transistors, operational amplifiers, and the like, do not satisfy Ohm’s law and are known as nonlinear devices (e.g., curve B in Figure 16.4). In fact, most if not all electronic devices have both resistors and nonlinear circuit elements in them.
Next we briefly consider the general topic of electrical energy and power. In the simple circuit of Figure 16.2, the battery terminals are maintained at a constant potential difference by chemical energy with the positive terminal at potential Vbattery with respect to the negative terminal at V⫽ 0. When the wire of length L is con-nected between the terminals, an electric current flows from the positive to negative terminal. If we plot the electric potential as a function of position along the wire (Figure 16.7), we see that it decreases linearly from the battery voltage at the posi-tive terminal to zero at the negaposi-tive terminal of the battery. A (posiposi-tive) charge ⌬Q flowing from the positive to negative battery terminal flows down this potential hill so that the decrease in electric potential energy is
(16.8)
Because in a time ⌬t, the charge flowing in the wire is ⌬Q ⫽ I ⌬t, the rate at which electric energy is lost is given by the electric power P,
(16.9)
The SI unit for electric power is the Watt, just as for all other powers, as can be ver-ified by substituting units for IV, 1 A⫻ 1 V ⫽ 1 CV/s ⫽ 1 J/s ⫽ 1 W.
If we examine the flow of energy in this example, stored chemical energy of the battery is used to maintain a constant potential difference between the battery termi-nals. This constant V produces a constant E field within the wire that, in turn, main-tains a constant drift velocity for the charges. Thus, the kinetic energy of the charges remains constant along the wire, although energy is continually lost through colli-sions. As charge flows along the wire and down the potential hill of Figure 16.7, the potential energy loss at a rate P appears as thermal energy of the wire causing a tem-perature increase. This transfer of energy occurs through the collisions with the array of metal atoms in the wire while the drift velocity is maintained by the constant elec-tric field using energy supplied by the battery. The elecelec-trical energy is said to be lost because the entire process is irreversible. As we have seen in our study of thermody-namics, a loss of potential energy of any kind to heat cannot be a truly reversible process.
Other expressions can be obtained for the power in terms of the resistance of the wire in our example. Using Ohm’s law, Equation (16.7), to eliminate either V or I, we obtain (16.10) P⫽ IV ⫽ I2 R⫽V 2 R. P⫽ ¢PE E ¢t ⫽ ¢Q ¢t V⫽ IV. ¢PE E⫽ ¢Q V.
Corresponding values are then, for distilled water, R⫽ 0.1 ⫻ 1012 ⍀ and I ⫽ 1.0 nA; for fresh water, R⫽ 2.5 ⫻ 108⍀ and I ⫽ 0.4 A; and for sea water, R ⫽ 0.15 M⍀ and I ⫽ 0.67 mA. The huge increase in current of almost a factor of one million is due to the increase in ion content of the sea water versus fresh water versus distilled water.
R⫽ rL/A ⫽ r 0.1
p(0.05/2)2⫽ 51r.
Vbattery
V
distance along wire L
FIGURE 16.7The voltage, measured with respect to the negative terminal of the battery, along the uniform wire of length L in Figure 16.2.
This conversion of electrical energy to thermal energy in a resistor is known as Joule heating. It is beneficially used in devices such as toasters, electric ovens, and heaters, but is a major source of energy loss in most other electrical devices. Excess heating can also be a fire hazard in poorly designed or defective house electrical wiring.
Example 16.2Calculate the power consumption for the three situations in Example 16.1. Also, find the rate at which the water temperature increases if no heat is lost to the surroundings.
Solution:The power calculation is straightforward using, for example, V2/R, to find powers of 0.1W (distilled water), 40 W (fresh water), and 67 mW (sea water). If none of the input power is lost, it is all converted to heat in the water. The water temperature will rise at a rate determined from
where P is the I2R Joule heating. The volume of water is given by r2L⫽ (3.14)(.025)2(0.1)⫽ 2.0 ⫻ 10⫺4m3, so that the mass of the water is about 0.2 kg, roughly independent of the salt concentration. Using a specific heat of 4180 J/(kg°C), we find rates of temperature increase of 1.2⫻ 10⫺10°C/s (for dis-tilled water), 4.8⫻ 10⫺8°C/s (for fresh water), and 8.0⫻ 10⫺5°C/s (for sea water). These heating rates are quite negligible, taking several hours to heat the sea water 1°C. However, if the tube length is decreased by a factor of 10 and the tube diam-eter is increased by a factor of 10, then the resistance will decrease by a factor of 1000, and both the current and power will increase by that same factor. In this case the heating is appreciable, increasing the sea water temperature by about 5°C/min.
¢Q ¢t ⫽ mc
¢T ¢t ⫽ P,
2. OHM’S LAW APPLICATIONS AND ELECTRICAL
MEASUREMENTS
Now that we have learned about electric current and resistance as well as potential, in this section we learn how to measure these in actual circuits and how to analyze some basic circuits. There are three common types of electric meters, often packaged in a multipurpose device known as a multimeter. By flipping a switch this device can measure current (as an ammeter), voltage (as a voltmeter), or resistance (as an ohm-meter). Although today these devices consist of complex semiconductor components, the fundamental principles of the devices can be more simply explained. Given a sim-ple circuit consisting of a battery and resistor as shown in Figure 16.8, how can one use a multimeter to measure the current in the circuit, the voltages across the battery or resistor, and the resistance value of the resistor?
Voltmeter V A Ammeter R Ohmmeter R R a b c A B
FIGURE 16.8Measurement of (a) the current through R, with an ammeter inserted into the circuit in series with R; (b) the voltage across R, with a voltmeter in parallel with R; or (c) the value of the resistance R itself, with an ohmmeter after removing the resistor from the circuit, as shown above.
Any electrical measuring device has its own internal resistance that must be designed to minimize the impact of the presence of the meter on the electrical prop-erties being measured. To measure the current in the circuit of Figure 16.8a, the mul-timeter must be set to act as an ammeter and be inserted into the circuit by “breaking” a wire (actually by replacing the one wire between the resistor and bat-tery with two wires) and inserting the meter “in series” with the resistor. Being “in series” means that the same current must flow through the ammeter as flows through the resistor; there is no other path for the current to follow. However, the presence of the ammeter, with its internal resistance, affects the total resistance in the circuit and thereby the current. We would like to “analyze” this circuit; that is, we would like to write the equations that allow us to predict the current the ammeter would measure for given values of the battery voltage, resistance, and ammeter resistance. There is a very general method to analyze circuits, even very complex ones, known as Kirchoff ’s loop equation. In this analysis, starting at an arbitrary point in the circuit diagram, one mentally “travels” around a closed loop, adding and sub-tracting the potential increases and decreases algebraically as the loop is traversed. The sum must add to zero because on returning to the starting position, the potential has that same starting value and thus the potential difference around any closed loop must be zero. In using the loop method, care is needed in choosing the proper alge-braic sign for the potential difference across each circuit element. For batteries the potential increases when going from the⫺ to ⫹ terminal across the device, whereas for resistors, the potential drops in going across the resistor in the direction of the cur-rent flow according to Ohm’s law. Whichever direction one chooses to go mentally around a loop, a consistent set of potential differences must be summed to zero for the loop method to work properly. Let’s continue with our analysis of Figure 16.8a; below we show the benefit of the loop equation in more complex circuit analysis.
Starting at the negative battery terminal (side with the shorter line in the symbol), we mentally “travel” around the loop clockwise (our arbitrary choice) adding and subtracting the appropriate voltages using Kirchhoff’s loop equation for circuit (a) in the figure to obtain
or
where
(16.11)
In this equation, V is positive because we are “traveling” from the⫺ to ⫹ terminal, and the IR voltages across resistors are both decreases (drops), taken as negative, because we are “traveling” around in the direction of the actual current flow from the ⫹ terminal of the battery. Our answer for this circuit is actually an example of a gen-eral result when any two (or more) resistors are connected in series:
The equivalent resistance of resistors in series is the sum of their individual resistances.
It also suggests that for an ammeter to have a negligible effect on the current in the original circuit, it must have a very small resistance, certainly negligible compared to the resistance in the circuit. Modern ammeters have a very low resistance, typically less than 1 ⍀. Given values for V and R, the equation above predicts the measured ammeter current.
In order to measure the voltage across any component in a circuit, a multimeter is set to act as a voltmeter and needs to have its terminals connected across that cir-cuit element as shown in Figure 16.8b to measure the voltage across the resistor. The voltmeter resistance is said to be “in parallel” with resistor R because both elements have the same potential difference across them. However, the current flowing out of
R equiv⫽ R ⫹ R ammeter.
(resistors in series) V⫽ IRequiv,
the positive terminal of the battery, when arriving at point A in the figure, divides with part of the current flowing through each “branch” of the circuit later to recombine at point B. This is our first example of a multiloop circuit, one in which the same current does not flow through all the circuit elements, and we digress further to show how it can be analyzed.
Consider the circuit shown on the left in Figure 16.9, similar to that of Figure 16.8b because the voltmeter is represented by a resistor in parallel with the original resis-tor. Using Kirchhoff’s loop equation to analyze this circuit, we can write down several equations depending on the chosen loop:
clockwise around the outer loop, starting from B; clockwise around the lower loop, from B; (16.12) clockwise around the upper loop, from B.
Clearly these equations are not all independent, because, for example, subtracting the second from the first results in the third. An additional independent equation can be obtained by noting that at points A and B (branch points) where the current divides, by conservation of electric charge we must have that
(16.13)
where I is the current from the battery (see Figure 16.9 left). This is an example of a second more general rule, known as Kirchoff ’s junction rule, which states that at a branch point (or junction) where several wires come together, the total current enter-ing the branch point must equal the total current leaventer-ing that point. Clearly this is a consequence of the general law of conservation of electric charge. Solving the first two equations of Equations (16.12) for each of the currents I1and I2and substituting into Equation (16.13), we can write that
where Requivis the single equivalent resistor that, when connected across the same battery voltage V will cause the same current I to flow from the battery (see the right side of Figure 16.9), so that V⫽ I Requiv. Dividing by V, we obtain
(16.14)
showing the general rule for resistors in parallel:
The inverse of the equivalent resistance of resistors in parallel is the sum of the inverses of individual resistances.
Returning to the measurement of the voltage across a resistor in the circuit of Figure 16.8b, by putting the voltmeter in parallel with the resistor the equivalent resistance seen by the battery will change (actually, it will always decrease; can you show this from Equation (16.14)?) and therefore so will the current flowing out of the battery (it will always increase in such a circuit). The excess current will be drawn into the voltmeter loop of the circuit. The battery current is entirely determined by the “load”, or equivalent resistance, on the battery from V⫽ IRequiv. To avoid chang-ing the battery current significantly, the voltmeter must have a very high resistance, so that it draws negligible current and the equivalent resistance is essentially that of the circuit, R. Modern voltmeters have resistances of about 10 M⍀ (1 M⍀ ⫽ 106⍀).
1 Requiv⫽ 1 R1⫹ 1 R2,
(resistors in parallel) I⫽ V Requiv⫽ V R1⫹ V R2, I⫽ I1 + I 2, I2R2⫺ I1 R1⫽ 0, V⫺ I2 R2⫽ 0, V⫺ I1 R1⫽ 0, R1 R2 V I1 I2 V Requiv I I B A
FIGURE 16.9A simple circuit with two resistors in parallel.
A multimeter can also function as an ohmmeter when directly connected to both sides of (across) a resistor which has been removed from the circuit, as in Figure 16.8c. By using an internal battery to send a known current through the resistor and by measuring the voltage across the resistor, the ohmmeter directly measures its resistance.
Example 16.3Find the current that flows through each of the resistors shown in the circuit of Figure 16.10. Also determine the power generated in each resistor.
Solution:In solving circuit analysis prob-lems it is important to first take a careful look at the “lay of the land” or the circuit’s basic “topology.” In this example, the 12 V battery is the only source of current in the circuit and so it sends current out of its⫹ terminal that then divides at the lower left branch point, some traveling through the 5 k⍀ resistor and the rest traveling through the 1.5 k⍀ and 2.5 k⍀ resistors, which are in series with each other. The currents in these two branches (the 5 k⍀ branch and
the (1.5 k⍀ ⫹ 2.5 k⍀) ⫽ 4 k⍀ branch) recombine in the upper right corner and their sum, the net battery current, then travels through the 2 k⍀ resistor and returns to the⫺ terminal of the battery. It is very important for you to be able to understand and eventually generate this type of qualitative analysis before going to equations in order to find values for the currents.
With the understanding of the previous paragraph, we can solve this problem in a simple straightforward manner, by finding the total equivalent resistance in the circuit from the following: (1) first, the 1.5 k⍀ and 2.5 k⍀ are in series and together have a net resistance of 4 k⍀ shown on the left below; (2) then the 4 k⍀ and 5 k⍀ are in parallel with each other (do you see why?), so that their equiva-lent resistance R is given by 1/R⫽ 1/4k⍀ ⫹ 1/5k⍀, giving R ⫽ 2.22 k⍀, shown in the middle below; (3) then the 2.22 k⍀ and the 2 k⍀ are in series with each other yielding a net resistance in the circuit of 4.22 k⍀, shown on the right.
The circuit on the right tells us that the current out of the battery is just I⫽ (12 V/4.22 k⍀) ⫽ 2.84 ⫻ 10⫺3A⫽ 2.84 mA. All of this current passes through the 2 k⍀ resistor because it is in series with the battery, but each of the other resis-tors only gets part of this current. To find how the current divides, we can work backwards in the set of figures just above. The current divides at the branch point so that the voltages across the 5 k⍀ and equivalent 4k resistor (see the left figure above) are equal because the two branch points have a fixed potential V between them whether we “travel” through the 5 k⍀ or 4 k⍀ resistor. This implies that
V⫽ (I5 k 5 kÆ)⫽ (I4 k 4 kÆ) 12V 2KΩ 5KΩ 1.5KΩ 2.5KΩ 12V 2kΩ 5kΩ 4kΩ 12V 2kΩ 2.22kΩ 12V 4.22kΩ
FIGURE 16.10Circuit for Example 16.3. Which resistors are in series or parallel with the others?
The preceding example was solved by simply using the rules for combining various resistors in series and parallel. There are more complex circuits where this type of analysis is not possible and Kirchoff’s loop equation must be used. The next example has such a circuit.
so that I5k/I4k⫽ 4/5. But we know that the total current, I5k⫹ I4k, is 2.84 mA, so that we can find the individual currents from either the previous two equations with their two unknown currents, or from the following simple argument. By dividing the total current in 9 parts (based on the ratio equation above using 9⫽ 4⫹ 5) we note that (4/9) of the total current, or 1.26 mA, flows through the 5 k⍀ and (5/9) of the total current, or 1.58 mA, flows through the 4 k⍀ equivalent resistor. Finally returning to the original circuit, each of the 1.5 k⍀ and 2.5 k⍀ resistors have I4k⫽ 1.58 mA flowing through them. You should check that these results are consistent and add up properly; follow each current around the orig-inal circuit and check Kirchoff’s junction rule.
We finish this problem by noting that the power generated in each resistor is given by P⫽ I2R, so that if we know the values of the currents and resistors we
can simply compute these values to be P2k⫽ 0.016 W, P5k⫽ 0.0079 W, P1.5k⫽ 0.0037 W, and P2.5k⫽ 0.0062 W. Note that the power supplied by the battery, given by P⫽ ItotalV⫽ 0.034 W is equal to the total power dissipated in all the resistors. Check this yourself !
Example 16.4Find the current flowing though each resistor of the following circuit.
Solution:In this case, because of the sec-ond battery in the circuit we cannot simply combine resistors in series and parallel but must use Kirchoff’s loop equation. Using the set of labeled currents, which can be chosen arbitrarily as long as they are consistent, we can write down two loop equations to allow us to solve for the two unknown currents labeled I1and I2in the figure. We have already implicitly used the junction equation in choosing the sum of the two currents from the batteries as the
current in the central branch of the circuit. Follow the currents to the right junction point and check that they are self-consistent there as well. We need only choose two of the three possible loops: the top, bottom, or outer loops, but for practice we write all three down and then only use two of them to solve for I1and I2.
First around the outer loop, starting arbitrarily at the lower left corner and going clockwise, we have
Make sure you understand why the signs are as they are (these are not arbitrary). Around the top loop, starting at the upper left corner and still going clockwise (note: the direction is arbitrary, but it is perhaps a good idea always to “travel” around loops the same way to help reduce mistakes)
⫺12 V ⫹ I1 (2 kÆ)⫹ (I1⫹ I2)(1 kÆ)⫽ 0. ⫺12 V ⫹ I1(2 kÆ)⫺ I2(3 kÆ)⫹ 6 V ⫽ 0. I1 I2 12V 2kΩ 3kΩ 1kΩ 6 V I1+I2
FIGURE 16.11Multiloop circuit for Example 16.4. Do you see why these resistors are not in series or parallel with each other?
We can also consider simple electrical circuits that have two capacitors C1and C2connected either in series or in parallel to a battery as shown in Figure 16.12. As just studied in the case of resistors, there will be a single equivalent capacitor that, when connected to the same battery, will produce the same resulting final state: the same charge will flow from the battery, storing the same amount of potential energy as in the original situation with two capacitors. In the next section we show the effects of having both resistors and capacitors in the same circuit, but first we com-plete this section by calculating the equivalent capacitance corresponding to those equations for the equivalent resistance of series and parallel resistor combinations, Equations (16.11) and (16.14).
Consider the case of two capacitors in series as shown on the left in Figure 16.12. Using the fact that the voltage across a capacitor is proportional to the charge on it, we have that V1⫽ Q1/C1 and V2⫽ Q2/C2, where the charges are those on each capacitor. Now, consider the portion of the circuit outlined in the dotted lines. This section of the circuit is completely isolated electrically and if it was originally neu-tral must remain so. Therefore the net negative charge on the right plate of C1and the net positive charge on the left plate of C2 must add to zero, proving that Q1⫽ Q2. Then, using the loop equation, the voltage V across the battery is equal to the sum of the voltages V1and V2across each capacitor and we have that
(16.15)
where Q is the common charge on each capacitor. The battery supplies positive charge Q to the left plate of C1which then induces an equal negative charge on its adjoining right plate, resulting in an equal and opposite positive charge at the left plate of C2and an induced equal negative charge on its right plate. We show in the next section that this “charging” of the capacitors when first connected to a battery takes some finite time, depending on the stray electrical resistance of the circuit. Finally, we see that if we replace the two capacitors by a single equivalent capacitor with capacitance C, that in order to have the same charge stored on this capacitor we require that
(16.16)
Capacitors in series combine reciprocally, just as resistors in parallel do according to Equation (16.14).
Using a similar analysis for capacitors in parallel, we see from the right-hand portion of Figure 16.12 that we now have that the total charge Q supplied by the battery is the sum of the charges on both capacitors: Q⫽ Q1⫹ Q2. From this, we can write
(16.17) Q⫽ Q1⫹ Q2⫽ C1 V1⫹ C2 V2, V⫽Q C⫽ Q C1⫹ Q C2 or 1 C⫽ 1 C1⫹ 1 C2.
(capacitors in series) V⫽ V1⫹ V2⫽ Q C1⫹ Q C2,
Finally, although not needed, around the bottom loop, again clockwise from the lower left corner,
Now, picking any two of these three equations, we need to do the algebra to solve for the two unknowns. We find that I1⫽ 3.82 mA and I2⫽ 0.55 mA. Check this for yourself.
⫺ ( I1⫹ I2)(1 kÆ)⫺ I2(3 kÆ)⫹ 6 V ⫽ 0. V C1 C2 V1 V2 V C2 C1
FIGURE 16.12Two capacitors in a (left) simple series or (right) parallel combination.
and again replacing the two capacitors with a single capacitor C and noting that the voltages across each capacitor are the same because they are in paral-lel (V1⫽ V2⫽ V), we find
(16.18)
Remember that, just as for resistors, these results for combining two capacitors in series or parallel can easily be generalized to larger arrays of capacitors using the same tools as in the above discussion. Circuits with only resistors or only capacitors present are ideals. In the next section we turn to a presentation of more realistic circuits with both resistors and capacitors present. Such circuits are more realistic because there is always a small amount of resistance (in the conducting wires themselves) or stray capacitance (between different conducting surfaces) present in any circuit regardless of whether an actual resistor or capacitor device is present in the circuit. We approach this topic using a model for cell membranes.
3. MEMBRANE ELECTRICAL CURRENTS
In the last chapter membranes were considered as ideal capacitors with a specific capacitance (capacitance per unit area) of about 1F/cm2.
This turns out to be a very good approximation for a pure phospholipid bilayer which has an extremely high resistivity of about 1015 ⍀-cm,
comparable to a very good insulator. The very high equivalent resis-tance prevents charge from crossing the lipid region and maintains the stored charge as if the bilayer were an ideal capacitor. However, as dis-cussed in the last chapter, biological membranes are full of proteins that act as channels allowing ionic currents to flow across a membrane.
The simplest model, or equivalent circuit, for a biological mem-brane in the resting state is shown in Figure 16.13 and is known as an RC series circuit. For now, we ignore how the equivalent capacitor was charged (to a voltage V0⫽ Q0/C) and we imagine that at time zero the switch S is closed (corresponding to the membrane channels opening), discharging the capacitor. The capacitor does not discharge instanta-neously, but follows a time course that depends on the values of R and C. The resistance R represents the effective resistance to current flow across the membrane and is discussed further below.
To analyze this circuit, we use Kirchhoff’s loop method, discussed in the last section. Let’s write a loop equation for the circuit in Figure 16.13 after the switch is closed and a path is provided for current flow. When the switch is closed current will flow from the ⫹Q0side of the capacitor clockwise around the circuit. Starting at the switch S and mentally going clockwise around the loop, we find
(16.19)
Because both Q and I vary with time, it turns out that we need calculus to solve this equation (see box) to find that the charge on the capacitor and the current through the resistor are given by
(16.20a) (16.20b) I⫽ I0 e⫺ t RC, Q⫽ Q0 e⫺ t RC, ⫺ IR ⫹Q C⫽ 0. C⫽ C1⫹ C2.
(capacitors in parallel) Q⫽ CV ⫽ C1 V⫹ C2 V or R C Qo –Qo S
Starting from Equation (16.19), and substi-tuting from the definition of
(the minus sign is needed to make the cur-rent positive because it is equal to the time rate of decrease of the capacitor charge), the equation becomes
.
Rewriting, we have
.
Integrating both sides of this equation from t⫽ 0 to time t and from Q(t ⫽ 0) ⫽ Q0to a value of Q(t), written simply as Q, we find
,
so that
.
Taking the antilog of both sides, remembering that these logarithms are to the base e, we find
,
or Equation (16.20a). To then find the cur-rent as a function of time, we again use its definition, so that
,
or Equation (16.20b). This same procedure can be used to analyze any electrical circuit consisting of batteries, capacitors, and resistors via the loop equation.
I⫽ ⫺dQ dt ⫽⫺ Q0 d(e⫺ t RC) dt ⫽ Q0 RCe ⫺ t RC Q Q0⫽ e ⫺t RC log Q⫺ log Q0⫽ logQ
Q0⫽⫺ t RC LQ 0 Q dQ Q ⫽⫺ 1 RCL t 0 dt dQ Q ⫽⫺ dt RC RdQ dt ⫹ Q C ⫽ 0 I⫽⫺dQ dt
FIGURE 16.13An RC series circuit with the capacitor initially charged before closing the switch S connected to the resistor.
where Q0 is the initial charge on the capacitor and I0 is the initial current when the switch is closed and given by I0⫽ Q0/RC.
The results obtained in Equations (16.20) are shown in Figure 16.14 with the charge and current plotted as functions of time. Because the voltage across the capaci-tor is proportional to the charge (V⫽ Q/C) and the volt-age across the resistor is also proportional to the current (V⫽ IR), these voltages follow the same time courses as Q and I, respectively. The key parameter in these results is
the product RC, which has units of time and is known as the RC time constant⫽ RC. Its value determines the rate at which the discharging of the capacitor occurs, with the charge, current, or voltage across either R or C dropping to (1/e)⫽ 0.37 of its initial value in a time ⫽ RC (see Figure 16.14).
All electrical devices and complete circuits have some associated capacitance as well as resistance. In high-speed electrical applications, such as computers, the RC time constant sets fundamental limits on the speed at which a circuit can change its voltage. Computers use voltage as information, with a high or low voltage represent-ing a bit of information, either a 1 or a 0, and calculations are done by electronic arithmetic that changes bits rapidly. Consequently, increasing the processing speed of a computer depends heavily on reducing the associated capacitance of the funda-mental electronic device building blocks of the microprocessor.
0 0.2 0.4 0.6 0.8 1 0 10 20 30 40 50 60 time (ms) Normalized Q or I
FIGURE 16.14Normalized capaci-tor charge or electric current in an RC circuit (Equations (16.20a) and (16.20b), normalized to their initial values) for a ⫽ 15 ms RC time constant. The voltages across the capacitor and resistor also follow the same time course. Dashed lines indicate that at t⫽, the normal-ized Q or I has decreased to (1/e)⫽ 0.37 of its starting value of 1.0.
Example 16.5In the simple RC circuit of Figure 16.13, the 10F capacitor is initially charged to 60C. When the switch is closed, an initial current of 0.3 mA is measured in the circuit. Find the charge on the capacitor and the current in the circuit after 0.6 s.
Solution:To learn the time course of the current and charge, we need to first find the value of the resistance in the circuit. When the switch is first closed, the ini-tial voltage across the resistor is the full iniini-tial voltage V0across the capacitor. Because the initial charge on the capacitor is 60C, the initial voltage is V0⫽ Q0/C⫽ 60 C/10 F ⫽ 6 V. This voltage, on closing the switch, immediately produces the given initial current flow I0⫽ 0.3 mA. From Ohm’s law R ⫽ V/I, so knowing the initial current we can solve for R⫽ (6 V)/(0.0003 A) ⫽ 20 k⍀. Now, knowing RC⫽ (20 k⍀)(10 F) ⫽ 0.2 s, we can use Equations (16.20) to find the charge and current after 0.6 s, equal to three time constants. Substituting that (t/RC)⫽ 3, we find that the exponential is given by e⫺3⫽ 0.05, so that after 0.6 s there will remain only 0.05 times the initial charge and current. Our answers then are that after 0.6 s there remain (60C)(0.05) ⫽ 3 C of charge and the current is (0.3 mA)(0.05)⫽ 15 A.
Let’s now apply some of these ideas to a biological membrane where we are par-ticularly interested in the transverse currents across the membrane. For membranes in the resting state, RC time constants range from 10s to 1 s. In dealing with mem-branes it is useful to discuss the electrical properties of a 1 cm2area; these are known
as the specific capacitance C/A, and specific resistance RA. Defined in this way the product of the specific capacitance and specific resistance (C/A)(RA)⫽ RC is still equal to the time constant. From R⫽L/A, we have that RA⫽L in units of ⍀-cm2.
Using the value quoted for the membrane specific capacitance C/A, in the previous chapter of 1F/cm2, the different time constants correspond to different values for
the specific resistance RA⫽L of 10 to 106⍀-cm2. The broad range of values for
the resistivity indicates a large variability in both the numbers of channels per unit area and in the average number of open channels in the resting state in different cells.
We now want to get some estimate of the numbers of charges flowing through each open channel that make up the membrane current. Using a value of 0.1 V for the resting potential, we determined in the last chapter that a typical value for the surface charge density Q0/A is about 0.1C/cm2. Because 1 mol of a monovalent ion
corre-sponds to a charge of , where F
is known as the Faraday constant, we can find the number of moles corresponding to a charge Q0per unit area. If due to monovalent ions, the surface charge density cor-responds to
If we approximate the average current density (current per unit area) by dividing the charge density value by the time constant, we find a current density I/A of 100A/cm2using a 1 ms time constant. This corresponds to the flow of 1 nmol of
ions/cm2/s. Using a value of about 10 channels/m2 (or 109 channels/cm2) for the
surface density of channels, the ratio of I/A (10⫺4 A/cm2) to channels/cm2 gives a
value for the current in a single channel of about 0.1 pA, corresponding to the flow of about 10⫺18mol of ions/s. This means that each channel carries about 600,000 ions/s or about 600 ions in the 1 ms time constant. Measured values for a variety of single channels give currents of this magnitude or 10–100 times larger (see Section 6). Note that the number of ions flowing across the membrane is insignificant in terms of the total concentrations of ions both in the cytoplasm and extracellular medium, so that the ion concentrations in these media remain essentially constant.
Thus far in our discussion we have ignored the membrane charging mechanism, or in the language of equivalent circuit diagrams, we have ignored a source of energy, a battery or power supply. What is the origin of the membrane resting potential? We show that the selective permeability of the membrane to various ions, controlled by the channels, is the source of this potential.
Suppose first that there are only K⫹channels in a membrane so that, to a good approximation, only those ions can cross the membrane barrier. If we start with an excess of KCl on one side of the membrane, the K⫹will reach an equilibrium across the membrane in which there is no net flow of ions even though the K⫹ concentra-tion is not equal on both sides of the membrane. Why is this? Clearly in the absence of any electrical effects, diffusion alone would tend to drive the K⫹concentration to the same final value on both sides of the membrane. However, despite this diffusional driving force, electrical attractive forces due to the presence of the excess (negative Cl⫺) ions, which cannot cross the membrane, balance this tendency toward a uniform concentration at equilibrium (Figure 16.15).
From an equilibrium equation similar to that of the discussion of Figure 13.6 in Chapter 13, we can write that
(16.21)
where R is the molar gas constant, and the c’s and PE’s are molar concentrations and potential energies, respectively, of the K⫹on the outside (o) and inside (i) of the membrane. Writing that
where NAis Avogadro’s number, z is the valence or number of charges per ion (so that is the charge of a mole of ions), and VKis the equilibrium membrane potential due to potassium ions. Solving for VKby taking the natural logarithm of Equation (16.21), we have (16.22) VK⫽ RT zF log a c0 cib. zF PE0 ⫺ PEi⫽ NAq¢V⫽ z F(V0⫺ V i)⫽ zFVK, c0 ci ⫽ e ⫺PE0RT⫺PEi , (10⫺7 C/cm2)/F⫽ 10⫺12 mol/cm2⫽ 1 p mol/cm2. F⫽ N A e⫽ 6 ⫻ 1023⫻ 1.6 ⫻ 10⫺19 L 105 C/mol FIGURE 16.15Portion of a membrane (with channels not shown) permeable only to K⫹(blue) showing that even at equilibrium, the concentration of K⫹is higher on the side with Cl⫺(pink) due to electrical forces.
The Nernst potential represents the equilibrium situation for a particular ion species. If the transmembrane potential is equal to the Nernst potential for some ion species “A,” VA, then there will be no net flow of A across the membrane even if the membrane has a high conductivity for A. No net flow does not mean that the chan-nels do not allow any ion flow, but rather that the inward and outward flows of ion A are equal. If the transmembrane potential is higher or lower than the Nernst potential then there will be a net flow of A one way or the other across the membrane with the ionic current proportional to the difference between the actual potential and the Nernst potential for that ion
(16.23)
where GAis the A ion conductance and V is the actual transmembrane potential. If only the one ion species can cross the membrane, then the membrane potential will equilibrate at the Nernst potential for that ion. In the resting state, open K⫹channels dominate and the resting potential is close to the equilibrium potential for K⫹, ⫺0.1 V. This behavior is identical to that expected if there were a battery in series with a resistor for each ion species. These separate batteries across the membrane function when their corresponding channels are open, corresponding to when their series resis-tance decreases.
At this point in our discussion we can present a more realistic circuit diagram for a membrane than a simple RC circuit. In the membranes of the axons of neu-rons, Na⫹ and K⫹ channels dominate, and Hodgkin and Huxley proposed the equivalent circuit shown in Figure 16.16. The arrows through the resistors in the figure indicate conductances that can vary with time as the ionic channels are made to open or close (known as gated channels). Only Na⫹and K⫹channels are explic-itly indicated with a net leakage conductance representing other net ion flows. Before we study some of the electrical properties of neurons and this equivalent circuit representation in Section 5, we first give a more qualitative overview of the structure and functioning of neurons and the ways in which their electrical properties have been studied.
4. OVERVIEW OF NERVE STRUCTURE AND FUNCTION;
MEASUREMENT TECHNIQUES
The human nervous system consists of some 1011nerve cells, or neurons, each one making an average of over 1000 interconnections. On an individual level we have a reasonable understanding of the functioning of a single nerve cell,
IA⫽ GA (V⫺ VA ),
Table 16.2Typical Ion Concentrations and Nernst Potentials (Mammalian Skeletal Muscle) Typical Internal Typical External Nernst Potential Ion Concentration (mM)* Concentration (mM) (mV)
Na⫹ 12 145 ⫹67 K⫹ 155 4 ⫺98 Ca2⫹ 10⫺4 1.5 ⫹129 Cl⫺ 4 120 ⫺90 * 1 mM⫽ 10⫺3M⫽ 10⫺3mol/L. IC IK IL outside inside C VNa VK VL GNa GK GL INa
FIGURE 16.16The Hodgkin– Huxley equivalent circuit for an axon membrane. The batteries represent the specific ion Nernst potentials (L⫽⫽ leakage, represent-ing the small contribution from other ions), producing specific ion currents as shown. The total membrane current is given by the sum of the four currents listed with the capacitor current equal to (from Q⫽⫽ CV)
where V is the voltage across the membrane.
IC⫽ C¢V ¢t ,
Equation (16.22) is known as the Nernst equation and determines the equilibrium membrane potential contribution from the imbalance of a particular ion, known as the Nernst potential. Table 16.2 gives typical concentrations and Nernst potentials for Na⫹, K⫹, Ca2⫹, and Cl⫺.
but we have precious little knowledge of the larger-scale, or more global func-tioning, of our nervous system. Three main ways to categorize nerve cells include whether they are part of the central (brain⫹ spinal cord) or peripheral (all else) nervous systems, part of the autonomic (connections with involuntary muscles and internal organs) or somatic (peripheral connections to voluntary muscles and sur-face sensors) nervous systems, or whether they are afferent (so-called sensory neurons, carrying information from the peripheral to the central nervous system) or efferent (so-called motor neurons, carrying information in the opposite direc-tion). There are many different types of neurons, however, they all have common features and are believed to function in a very similar manner.
Neurons are single cells with a cell body containing a nucleus and usually a single long thin structure, the axon, which may be more than 1 m in length. There are also several shorter processes, known as the dendrites, radiating away from the cell body (Figure 16.17). Cell bodies tend to be clustered together in regions connected by bundles of axons. At the far end of the axon are the termi-nal endings.
Nerve cells conduct an electrical signal called the action potential, or nerve impulse, discussed in detail in the next section. These signals are very similar in all nerves, traveling from the dendritic end to the terminal bundle end at speeds of up
Cell body axon Myelin sheath dendrites Terminal endings
FIGURE 16.17Structure of the neuron (top) schematic; bottom multiphoton scanning microscopy view of nerve bundles (green) and a retinal “starburst” cell (red) found in visual processing network.
to 100 m/s. Usually each neuron is electrically isolated from the next and signals are passed on to the next cell chemically. This occurs through the release of a neuro-transmitter from synaptic vesicles at the terminal endings. These chemicals diffuse across the synapse, a small cleft between the terminal endings of one neuron and the dendrites of the next, and are detected by membrane receptors on the dendrites to provoke an electrical response. Receptors are membrane bound proteins that, on binding neurotransmitters either directly (through so-called ligand-gated channels) or indirectly through open ion channels, cause a membrane depolarization and a continuation of the action potential. In certain neurons direct electrical connections between neighboring cells occur via “gap junctions,” pores connecting two neigh-boring cells that allow the direct passage of very small molecules. These are com-monly found in embryo tissue and are believed to provide a means for cell–cell communication in undeveloped tissue. In nerve cells, however, gap junctions do not allow as great a variety of control mechanisms as chemical synapses do, and are therefore relatively rare.
It is useful to describe the overall circuitry involved in a simple reflex response. At a minimum such a response requires four cells. The knee jerk reflex is well known as a simple reflex involving a muscle fiber, a receptor transducer cell, a sensory neuron, and a motor neuron. When a doctor taps the patellar tendon near the knee, the attached muscle is stretched. A stretch receptor senses this and produces an electrical response that is carried by an action potential along a sensory neuron to the spinal cord. There a reflex response is generated as an action potential in a motor neuron returning to the same muscle fiber. Arrival of this action potential generates a sequence of chemical steps that result in the contraction of the muscle, and the knee jerk response. A similar sequence of events occurs when you respond to a pinprick on your finger (Figure 16.18). Of course this is a simplistic view, and there are other neural connections that allow control over the sensory and motor signals from the central nervous system as well, but it serves to give a picture of the overall circuitry in a simple reflex.
Electrical properties of individual neurons can be studied in living tissue using inserted microelectrodes. Most of the early research work was done using the giant axon from a squid, a particularly large cell with an axon of about 1 mm in diameter. The electrode is a glass capillary tube containing a conducting salt solu-tion and a metal wire electrode. Electrodes are used both to measure membrane voltages (with the wire inside the tube connected to a sensitive voltmeter) and to inject small amounts of current (with the wire attached to a power supply). Usually the microelectrode is set to zero potential in the extracellular medium and, when inserted through the membrane into the cell, reads the resting membrane potential, typically a small (0.1 V) negative voltage with respect to the outside. When used to study a nerve impulse, often current is applied through a second electrode as a stimulus and subsequent changes in potential are measured. Alternatively, a constant voltage step change could be applied, fixing the mem-brane potential, and the changes in current flow across the memmem-brane measured. This method is known as the voltage-clamp technique.
On first thought, one might guess that the membrane could be voltage-clamped by connecting an ideal battery across its thickness. The battery would supply whatever current was needed to offset the membrane currents in order to maintain a fixed membrane potential. This is, how-ever, not quite true because the battery terminals cannot be “attached” to the membrane and there are unpredictable junction potentials at the metal–solution boundary due to contact resistance that would vary with the current flow. Only the metal electrodes would be voltage-clamped, not the membrane itself. Instead, voltage-clamping involves using an electric feedback loop to continually inject small currents in order to maintain a fixed potential.
Withdraw finger response motor neuron
sensory neuron
interneurons pin prick stimulus
Figure 16.19 shows three examples of voltage-clamp circuitry using feedback loops. In each method, the membrane potentials are “space-clamped” in such a way as to have no spatial variation of potential. In two of these methods two elec-trodes are used, with one measuring the potential relative to a reference voltage set at the desired level. This voltage difference signal is then used to inject a current through the second electrode to reduce the difference signal and maintain the volt-age clamp. Such a procedure is an example of negative feedback, in which an “error signal” is sent back to the source and used to make small corrections so as to restore a desired value of a variable. The space-clamping is achieved by either using long intracellular electrodes or by using a small membrane area isolated by either applying insulators in gaps dividing the membrane or by a patch-clamp arrangement. Patch-clamping, developed in 1976, uses a micron-diameter pipette tip pressed against an intact cell with some suction applied to form a very tight seal on a microscopic area of membrane so that the resistance between the inside and outside solutions is many G⍀ (1 G⍀ ⫽ 109 ⍀). Patch-clamping has led to a
100-fold increase in the sensitivity of membrane current measurements (see Section 6 below).
5. ELECTRICAL PROPERTIES OF NEURONS
When several electrodes are used to probe the spatial pattern of normal membrane potentials it is found that small cells have membrane electric potentials that are con-stant over their entire surface whereas larger cells, such as neurons, can have poten-tials that vary spatially as well as temporally. Although a small cell’s membrane can be reasonably modeled by a simple single-loop circuit diagram, Figure 16.20, in which the membrane voltage and current values depend on time, but not on spatial location (a so-called lumped-parameter model), neurons cannot.
Modeling the electrical properties of a neuron requires a so-called distributed-parameter network. The simplest scheme for a neuron that leads to some useful results is a linear cable model shown in Figure 16.21. This ribbon of repeated circuit elements is characterized by a set of parameters that vary along the length x. Here the inner and outer conductors represent the intracellu-lar and extracelluintracellu-lar fluid. Each section of length ⌬x along the cable has per-unit-length values of membrane capacitance cM, conductivity gM, transverse (inner to outer) current Im, and inner and outer longi-tudinal resistance riand r0, as well as inner and outer values for lon-gitudinal current along the axon Ii and I0, and voltage difference across the membrane VM. The model was first developed to represent an electrical cable (hence the name) that leaks some current trans-versely across the insulation between the two co-axial conductors. Although the mathematics of this model is complex, it is based on a
ammeter injected current feedback amplifier input measured voltage feedback amplifier A measured voltage injected current ammeter input input measured voltage ammeter membrane patch pipet tip feedback amplifier A
FIGURE 16.19 Three types of voltage-clamps. From left to right: gap method with insu-lating dividers, double electrode method for cells, patch-clamp method for pieces (patches) of membrane. C G V Vmembrane Imembrane
FIGURE 16.20 Equivalent circuit for the membrane of a small cell with no spatial variation in its electrical parameters.
straightforward application of Kirchhoff’s rules. Here we are content with showing a few of the model’s predictions.
Two parameters of the model are needed: the RC (⫽ C/G) time constant, given by (16.24)
and the space constant given by
(16.25)
Note how the units work out in Equation (16.24), with both cMand gMper-unit-length constants so that their ratio has time units, whereas in Equation (16.25) the per-unit-length constants combine to give units of distance. The time constant is a property solely of the membrane with typical values of several ms, whereas the space constant depends also on the cell dimensions and geometry and has typical values of several mm. If a steady electric current is applied at one point (x⫽ 0) along a neuron, the mem-brane voltage difference VMfrom the resting potential decreases exponentially along the axon in either direction according to
(16.26)
as shown in Figure 16.22. After a brief initial time when the current is applied, this result is time-independent because current is continually injected by the electrode to achieve a steady state.
If, on the other hand, a short pulse of current is injected into an axon at x⫽ 0 at time zero, the model can be used to cal-culate the voltage response as a function of both position and time. This situation corresponds to a typical stimulation of a nerve or muscle membrane. Results for this model are plotted in two ways in Figure 16.23. On the left the spatial variation of the voltage response is shown for several different times (dif-ferent curves). At increasing times the response spreads out from x⫽ 0, decreasing in amplitude at x ⫽ 0, but increasing in amplitude at other locations for a brief time. This is perhaps better shown in the figure on the right where the time-dependence is plotted at several different distances from x⫽ 0 (given in units of ). The voltage rises and then falls with an exponential tail. The peak can be seen to move to farther loca-tions at later times, but with a rapidly decreasing amplitude. If
VM⫽ VM (0)e⫺ |x| l, l⫽ 1
1
(ri⫹ r0)gM . tM⫽cM gM, cΔx gΔx cΔx gΔx cΔx gΔx V V V Ii (x) Im (x) VM (x,t) VM (x+Δx,t) Io (x) riΔ(x) roΔx roΔx riΔ(x) Ii (x+Δx) Inside Outside Im (x+Δx) Io(x+Δx) x+Δx x+2Δx Ii (x+2Δx) Io(x+2Δx) Im (x+2Δx)FIGURE 16.21A cable model for the electrical properties of the membrane of a nerve axon. There are two parallel conductors along the inner and outer surfaces with repeated transmembrane circuit elements representing the local current-voltage characteristics that vary with position.
FIGURE 16.22The spatial variation in the membrane voltage from measurements along axons stimu-lated by a small current from an electrode at x⫽ 0.
the potential changes are below a threshold value, this will be the only response of the membrane, a localized brief signal. Data on so-called miniature end-plate poten-tials, due to spontaneously released neurotransmitters, are accurately modeled by the cable model. On the other hand, if the potentials exceed a threshold value, then a totally different type of behavior is observed: a nonlinear nerve pulse is initiated.
A nerve pulse, or action potential, is an all-or-nothing propagating potential wave that is the basis of all neural communication. The Hodgkin–Huxley (H-H) model is a generalization of the cable model in which the cross-membrane elements of the cable are spelled out in detail. In place of a single conductance channel, H-H uses three such paths, for K⫹, Na⫹, and for other leakage currents, with the conductances for Na⫹and K⫹given as variable conductances (shown with arrows through their equivalent resis-tor values in Figure 16.16). This latter change makes the entire problem nonlinear because the conductances for Na⫹ and K⫹ are now themselves functions of both membrane voltage and time. From Equation (16.23), we see that the ionic currents will now depend on the membrane voltage in some nonlinear way (with the exponent of VMnot equal to 1).
The crux of the H-H model is the specification of the conductances GNaand GK. Hodgkin and Huxley obtained these functions by fitting data from space-clamped mea-surements (eliminating the x-dependence, or the cable properties) that were also volt-age-clamped, allowing direct measurement of membrane currents. Individual membrane currents due to Na⫹ and K⫹ were measured by a number of methods, including radioactive labeling of the salt ions, or using channel blockers, specific chem-icals that block, or shut off, only one type of ion channel. From numerous measure-ments of currents at specific membrane voltages, plots of the conductances of each type of channel as functions of potential were obtained. With empirical equations for these conductances, the H-H model can account for all of the features of an action potential. Figure 16.24 shows the time-dependence of an action potential and the asso-ciated ionic conductances. The Na⫹ conductance increases after a time delay relative to the potential, peaks with the potential, and then falls off more rapidly. Again relative to the potential, the K⫹ conductance rises more slowly and peaks after the fall of the potential. Although the H-H model was developed under space and voltage-clamped conditions, it can explain a large number of distinguishing features of an action potential, including: (1) an all-or-nothing response, with a threshold value of membrane current, in which a fixed pulse shape propagates down an axon at a constant speed; (2) an absolute refractory period of time after the action poten-tial during which a second action potenpoten-tial cannot be elicited; (3) a relative refractory period of time during which a second action potential can only be elicited by an elevated current level substantially beyond a lower threshold; (4) a specific strength-duration relation giving the threshold current for
0 0.2 0.4 0.6 0.8 1 1.2 –4 –2 0 2 4 x/λ normalized V m normalized V m t/τ=1/16 t/τ=1 0 0.2 0.4 0.6 0.8 1 0 1 2 3 4 t/τ
FIGURE 16.23(left) Spatial dependence of spreading membrane voltage at various times (decreasing voltage curves at x⫽ 0 correspond to t/ ⫽ 1/16, 1/8, 1/4, 1/2, and 1); (right) Time-dependence of membrane potential at various distances from the stimulus at x⫽ 0 (decreasing peak voltage curves are at x/ ⫽ 0.5, 0.75, 1.0, 1.5 and 2).
FIGURE 16.24Membrane voltage changes during an action potential (bold), together with sodium and potassium ion conductances across the membrane.
