Electric Current and Cell
Membranes
16
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 membrane. 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 quiescent 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 membranes. Membranes act as capacitors, storing charge and electric potential energy, but
because of “leakage” of charge through channels, membranes are not the ideal capacitors treated in the last chapter. In order to discuss more realistic models for membrane 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 connected, 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
I⫽
¢Q
,
(16.1)
¢t
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
J. Newman, Physics of the Life Sciences, DOI: 10.1007/978-0-387-77259-2_16,
© Springer Science+Business Media, LLC 2008
ELECTRIC CURRENT
AND
R E S I S TA N C E
401
–Q
+Q
E
I
FIGURE 16.1 Two charged
conducting plates connected
by a conducting wire at time zero.
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.
voltage
plates, the electric field that drives the electric charges decreases. If the initial charge
on each capacitor plate was 1 ␮C 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/
1 ␮s ⫽ 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, electrostatic fields cannot exist within a conductor. The electric field that drives the electric 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 electric 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 electrons 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 collisions 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 cancel 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 separation 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 106 m/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
wire
+
battery
battery
eE
F
⫽ ,
m
m
(16.2)
wire
Distance around circuit
from battery
402
a⫽
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).
ELECTRIC CURRENT
AND
CELL MEMBRANES
The mean time between collisions, ␶, is so short that the electrons only acquire a very slow
drift velocity of about 10⫺3 m/s given by
eE
eVt
.
t⫽
vdrift ⫽ at ⫽
m
mL
(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
I⫽
¢Q
¢t
⫽
nAle
⫽ nAevdrift.
(l/ vdrift)
A
vd
l
FIGURE 16.3 Free charge in a wire
of cross-sectional area A and
length l traveling with a drift
velocity vd.
(16.4)
Substituting from Equation (16.3), the electric current is
I⫽
ne2t A
V.
mL
(16.5)
Defining the conductivity ␴ of the wire, an intrinsic property of the material, to be
s⫽
ne2t
,
m
we can rewrite Equation (16.5) as
A
I ⫽ s V ⫽ GV,
L
(16.6)
where G is known as the conductance.
Solving for V, this can be rewritten in terms of the resistance R
V ⫽ IR,
(16.7)
where
R⫽
1 L
L
1
⫽
⫽r .
s
G
A
A
The resistivity of the material ␳ is given by the inverse of the conductivity,
r⫽
1
,
s
both intrinsic parameters. This definition is made in analogy with the equality
between the resistance and the inverse of the conductance
R⫽
1
,
G
except that both of these quantities are dependent on the size and shape of the material, 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 equivalent statement of Ohm’s law is that the resistivity of the material remains a constant,
independent of the applied voltage.
ELECTRIC CURRENT
AND
R E S I S TA N C E
A
B
Slope = 1/R
I
V
FIGURE 16.4 The I–V curve for
an ohmic circuit element (A) and a
semiconductor diode (B).
403
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 conductance, the reciprocal of resistance, 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.
FIGURE 16.5 An assortment
of resistors.
Table 16.1 Resistivities of Various Materials (20°C)
Material
Conductors
Aluminum
Copper
Iron
Mercury
Silver
Tungsten
R2
FIGURE 16.6 A simple circuit
diagram showing a battery
connected to two resistors, one
wired after the other.
404
2.8 ⫻ 10⫺8
1.7 ⫻ 10⫺8
10. ⫻ 10⫺8
96. ⫻ 10⫺8
1.6 ⫻ 10⫺8
5.6 ⫻ 10⫺8
Ionic materials
Water (distilled)
Fresh water
Sea water
Cytoplasm
Fatty tissue
~2 ⫻ 105
~5 ⫻ 102
~0.3
~0.5
~15
Semiconductors
Germanium
Silicon
~0.5
~2. ⫻ 103
Insulators
Air (dry)
Glass
Rubber
R1
Resistivity, ␳ (⍀ . m)
4 ⫻ 1013
1010 ⫺ 1014
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 difference 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 dimensions of the water tube, we find that
ELECTRIC CURRENT
AND
CELL MEMBRANES
R ⫽ rL/A ⫽ r
0.1
⫽ 51r.
p(0.05/2)2
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.
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 connected 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 positive terminal to zero at the negative terminal of the battery. A (positive) charge ⌬Q
flowing from the positive to negative battery terminal flows down this potential hill
so that the decrease in electric potential energy is
¢PEE ⫽ ¢Q V.
(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,
P⫽
¢PEE
¢t
⫽
¢Q
¢t
V ⫽ IV.
(16.9)
The SI unit for electric power is the Watt, just as for all other powers, as can be verified 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 terminals. This constant V produces a constant E field within the wire that, in turn, maintains 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 collisions. 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 temperature 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 electric field using energy supplied by the battery. The electrical energy is said to be lost
because the entire process is irreversible. As we have seen in our study of thermodynamics, 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
P ⫽ IV ⫽ I 2 R ⫽
ELECTRIC CURRENT
AND
R E S I S TA N C E
V2
.
R
(16.10)
V
Vbattery
distance along wire
L
FIGURE 16.7 The voltage,
measured with respect to the
negative terminal of the battery,
along the uniform wire of length L
in Figure 16.2.
405
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 16.2 Calculate 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 0.1 ␮W (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
¢Q
¢t
⫽ mc
¢T
⫽ P,
¢t
where P is the I2R Joule heating. The volume of water is given by ␲r2L ⫽
(3.14)(.025)2(0.1) ⫽ 2.0 ⫻ 10⫺4 m3, 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 distilled 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 diameter 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.
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 ohmmeter). Although today these devices consist of complex semiconductor components,
the fundamental principles of the devices can be more simply explained. Given a simple 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?
V
R
A
Voltmeter
Ohmmeter
R
B
R
Ammeter A
a
b
c
FIGURE 16.8 Measurement 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.
406
ELECTRIC CURRENT
AND
CELL MEMBRANES
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 properties being measured. To measure the current in the circuit of Figure 16.8a, the multimeter 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 battery 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 subtracting 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 algebraic 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 current 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
V ⫺ IR ⫺ IRammeter ⫽ 0,
or
V ⫽ IRequiv,
where
Requiv ⫽ R ⫹ R ammeter.
(resistors in series)
(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 general 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 circuit 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
O H M ’ S L AW A P P L I C AT I O N S
AND
ELECTRICAL MEASUREMENTS
407
A
I2
R2
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.
I
Consider the circuit shown on the left in Figure 16.9,
similar to that of Figure 16.8b because the voltmeter is
V
represented by a resistor in parallel with the original resistor. Using Kirchhoff’s loop equation to analyze this
circuit, we can write down several equations depending on the chosen loop:
Requiv
R1
I1
B
I
V
FIGURE 16.9 A simple circuit with
two resistors in parallel.
V ⫺ I1 R1 ⫽ 0, clockwise around the outer loop, starting from B;
V ⫺ I2 R 2 ⫽ 0, clockwise around the lower loop, from B;
(16.12)
I2R2 ⫺ I1 R1 ⫽ 0, 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
I ⫽ I1 + I2,
(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 entering the branch point must equal the total current leaving 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 I1 and I2 and substituting
into Equation (16.13), we can write that
I⫽
V
Requiv
⫽
V
V
⫹ ,
R1 R2
where Requiv is 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
1
Requiv
⫽
1
1
⫹ ,
R1 R2
(resistors in parallel)
(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 changing 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 ⍀).
408
ELECTRIC CURRENT
AND
CELL MEMBRANES
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 16.3 Find 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 prob12 V
2 KΩ
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
5 KΩ
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
1.5 KΩ
2.5 KΩ
5 k⍀ resistor and the rest traveling through FIGURE 16.10 Circuit for Example
the 1.5 k⍀ and 2.5 k⍀ resistors, which are 16.3. Which resistors are in series or
in series with each other. The currents in parallel with the others?
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 equivalent 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.
12 V
12 V
2 kΩ
5 kΩ
2 kΩ
12 V
4.22 kΩ
2.22 kΩ
4 kΩ
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⫺3 A ⫽ 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 resistors 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 ⫽ (I5k 5 kÆ) ⫽ (I4k 4 kÆ)
(Continued)
O H M ’ S L AW A P P L I C AT I O N S
AND
ELECTRICAL MEASUREMENTS
409
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 original 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 !
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.
Example 16.4 Find the current flowing though each resistor of the following circuit.
12 V
Solution: In this case, because of the sec2 kΩ
ond battery in the circuit we cannot simply
combine resistors in series and parallel but
I1
1 kΩ
must use Kirchoff’s loop equation. Using
the set of labeled currents, which can be
I1 + I2
I2
chosen arbitrarily as long as they are
consistent, we can write down two loop
3 kΩ
equations to allow us to solve for the two
6V
unknown currents labeled I1 and I2 in the FIGURE 16.11 Multiloop circuit for
figure. We have already implicitly used the Example 16.4. Do you see why these
junction equation in choosing the sum of resistors are not in series or parallel
the two currents from the batteries as the with each other?
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 I1 and I2.
First around the outer loop, starting arbitrarily at the lower left corner and
going clockwise, we have
⫺12 V ⫹ I1(2 kÆ) ⫺ I2(3 kÆ) ⫹ 6 V ⫽ 0.
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.
410
ELECTRIC CURRENT
AND
CELL MEMBRANES
Finally, although not needed, around the bottom loop, again clockwise from the
lower left corner,
⫺ (I1 ⫹ I2)(1 kÆ) ⫺ I2(3 kÆ) ⫹ 6 V ⫽ 0.
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.
We can also consider simple electrical circuits that have two capacitors C1 and
C2 connected 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 complete 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 neutral must remain so. Therefore the net negative charge on the right plate of C1 and 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 V1 and V2 across each capacitor and we have that
V ⫽ V1 ⫹ V2 ⫽
Q
C1
⫹
Q
C2
,
(16.15)
where Q is the common charge on each capacitor. The battery supplies positive
charge Q to the left plate of C1 which then induces an equal negative charge on its
adjoining right plate, resulting in an equal and opposite positive charge at the left
plate of C2 and 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
V⫽
Q
C
⫽
Q
C1
⫹
Q
C2
or
1
1
1
⫽
⫹ .
C C1 C2
(capacitors in series) (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
Q ⫽ Q1 ⫹ Q2 ⫽ C1V1 ⫹ C2V2,
O H M ’ S L AW A P P L I C AT I O N S
AND
ELECTRICAL MEASUREMENTS
FIGURE 16.12 Two capacitors in a
(left) simple series or (right) parallel
combination.
C1
C2
V1
V2
C1
C2
V
(16.17)
V
411
Starting from Equation (16.19), and substituting from the definition of
I ⫽⫺
dQ
Q ⫽ CV ⫽ C1V ⫹ C2V or
dt
(the minus sign is needed to make the current positive because it is equal to the time
rate of decrease of the capacitor charge), the
equation becomes
R
dQ
dt
⫹
Q
C
⫽ 0.
Rewriting, we have
dQ
Q
⫽⫺
dt
.
RC
Integrating both sides of this equation from
t ⫽ 0 to time t and from Q(t ⫽ 0) ⫽ Q0 to a
value of Q(t), written simply as Q, we find
Q
t
1
,
⫽⫺
dt
LQ0 Q
RC L0
dQ
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 parallel (V1 ⫽ V2 ⫽ V ), we find
C ⫽ C1 ⫹ C2.
(16.18)
(capacitors in parallel)
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 1 ␮F/cm2.
This turns out to be a very good approximation for a pure phospholipid
Q
t
bilayer which has an extremely high resistivity of about 1015 ⍀-cm,
log Q ⫺ log Q0 ⫽ log ⫽⫺
.
Q0
RC
comparable to a very good insulator. The very high equivalent resistance prevents charge from crossing the lipid region and maintains the
Taking the antilog of both sides, remembering
stored charge as if the bilayer were an ideal capacitor. However, as disthat these logarithms are to the base e, we find
cussed in the last chapter, biological membranes are full of proteins that
t
Q
⫺
act as channels allowing ionic currents to flow across a membrane.
RC
⫽e
,
The simplest model, or equivalent circuit, for a biological memQ0
brane in the resting state is shown in Figure 16.13 and is known as an
or Equation (16.20a). To then find the curRC series circuit. For now, we ignore how the equivalent capacitor was
rent as a function of time, we again use its
charged (to a voltage V0 ⫽ Q0/C) and we imagine that at time zero the
definition, so that
switch S is closed (corresponding to the membrane channels opening),
t
discharging the capacitor. The capacitor does not discharge instanta⫺
t
RC
dQ
d(e
) Q0 ⫺
neously, but follows a time course that depends on the values of R and C.
RC
⫽ ⫺ Q0
I ⫽⫺
,
⫽
e
dt
The resistance R represents the effective resistance to current flow
dt
RC
across the membrane and is discussed further below.
or Equation (16.20b). This same procedure
To analyze this circuit, we use Kirchhoff’s loop method, discussed
can be used to analyze any electrical circuit
in
the
last section. Let’s write a loop equation for the circuit in Figure
consisting of batteries, capacitors, and
16.13
after
the switch is closed and a path is provided for current flow.
resistors via the loop equation.
When the switch is closed current will flow from the ⫹Q0 side of the
capacitor clockwise around the circuit. Starting at the switch S and mentally going
clockwise around the loop, we find
so that
C
–Qo
Qo
S
R
FIGURE 16.13 An RC series circuit
with the capacitor initially charged
before closing the switch S
connected to the resistor.
412
Q
⫽ 0.
(16.19)
C
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
⫺ IR ⫹
Q ⫽ Q0 e
I ⫽ I0 e
⫺
⫺
t
RC
,
(16.20a)
t
RC
,
ELECTRIC CURRENT
(16.20b)
AND
CELL MEMBRANES
Normalized Q or I
1
where Q0 is the initial charge on the capacitor and I0 is
the initial current when the switch is closed and given by
0.8
I0 ⫽ Q0/RC.
0.6
The results obtained in Equations (16.20) are shown
0.4
in Figure 16.14 with the charge and current plotted as
functions of time. Because the voltage across the capaci0.2
tor is proportional to the charge (V ⫽ Q/C) and the volt0
age across the resistor is also proportional to the current
0
10
20
(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 representing 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 fundamental electronic device building blocks of the microprocessor.
30
time (ms)
40
50
60
FIGURE 16.14 Normalized capacitor 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 normalized Q or I has decreased to (1/e) ⫽
0.37 of its starting value of 1.0.
Example 16.5 In the simple RC circuit of Figure 16.13, the 10 ␮F capacitor
is initially charged to 60 ␮C. 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 initial voltage across the resistor is the full initial voltage V0 across the capacitor.
Because the initial charge on the capacitor is 60 ␮C, 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 (60 ␮C)(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 particularly interested in the transverse currents across the membrane. For membranes
in the resting state, RC time constants range from 10 ␮s to 1 s. In dealing with membranes it is useful to discuss the electrical properties of a 1 cm2 area; 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 1 ␮F/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.
MEMBRANE ELECTRICAL CURRENTS
413
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 0.1 ␮C/cm2. Because 1 mol of a monovalent ion corresponds to a charge of F ⫽ NA e ⫽ 6 ⫻ 1023 ⫻ 1.6 ⫻ 10 ⫺19 L 105 C/mol, where F
is known as the Faraday constant, we can find the number of moles corresponding to
a charge Q0 per unit area. If due to monovalent ions, the surface charge density corresponds to
(10 ⫺7 C/cm2)/F ⫽ 10 ⫺12 mol/cm2 ⫽ 1 pmol/cm2.
FIGURE 16.15 Portion 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.
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
100 ␮A/cm2 using 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⫺18 mol 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⫹ concentration 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
c0
⫺
⫽e
ci
PE0 ⫺PEi
RT
,
(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 PE0 ⫺ PEi ⫽ NAq¢V ⫽ z F(V0 ⫺ V i) ⫽ zFVK,
where NA is Avogadro’s number, z is the valence or number of charges per ion
(so that zF is the charge of a mole of ions), and VK is the equilibrium membrane
potential due to potassium ions. Solving for VK by taking the natural logarithm
of Equation (16.21), we have
VK ⫽
414
c0
RT
log a b.
ci
zF
ELECTRIC CURRENT
(16.22)
AND
CELL MEMBRANES
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⫺.
Table 16.2 Typical Ion Concentrations and Nernst Potentials (Mammalian Skeletal Muscle)
Ion
Typical Internal
Concentration (mM)*
Typical External
Concentration (mM)
Nernst Potential
(mV)
12
155
10⫺4
4
145
4
1.5
120
⫹67
⫺98
⫹129
⫺90
Na⫹
K⫹
Ca2⫹
Cl⫺
* 1 mM ⫽ 10⫺3 M ⫽ 10⫺3 mol/L.
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 channels 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
IA ⫽ GA (V ⫺ VA ),
(16.23)
where GA is 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 resistance 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 neurons, 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 explicitly 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.
FIGURE 16.16 The Hodgkin–
Huxley equivalent circuit for an
axon membrane. The batteries
represent the specific ion Nernst
potentials (L ⫽ leakage, representing 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)
IC ⫽ C
¢V
,
¢t
where V is the voltage across the
membrane.
outside
IC
4. OVERVIEW OF NERVE STRUCTURE AND FUNCTION;
MEASUREMENT TECHNIQUES
The human nervous system consists of some 1011 nerve 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,
O V E RV I E W
OF
N E RV E S T R U C T U R E
AND
FUNCTION; MEASUREMENT TECHNIQUES
INa
IK
GNa
C
VNa
VK
IL
GL
GK
VL
inside
415
but we have precious little knowledge of the larger-scale, or more global functioning, 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 surface 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 direction). 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 terminal 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
FIGURE 16.17 Structure 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.
dendrites
axon
Cell body
416
Myelin sheath
Terminal
endings
ELECTRIC CURRENT
AND
CELL MEMBRANES
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 neurotransmitter 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 neighboring cells that allow the direct passage of very small molecules. These are commonly 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 solution 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 membrane potential, and the changes in current flow across the membrane 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
FIGURE 16.18 A simple reflex circuit.
across its thickness. The battery would supply whatever
current was needed to offset the membrane currents in
pin prick stimulus
order to maintain a fixed membrane potential. This is, however, not quite true because the battery terminals cannot be
“attached” to the membrane and there are unpredictable
interneurons
junction potentials at the metal–solution boundary due to
sensory neuron
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
Withdraw finger response
using an electric feedback loop to continually inject small
motor neuron
currents in order to maintain a fixed potential.
O V E RV I E W
OF
N E RV E S T R U C T U R E
AND
FUNCTION; MEASUREMENT TECHNIQUES
417
measured
voltage
injected current
injected current
A
ammeter
input
membrane patch
measured
voltage
feedback
feedback
amplifier
measured
voltage
input
ammeter
feedback
amplifier
A
pipet tip input
amplifier
ammeter
FIGURE 16.19 Three types of voltage-clamps. From left to right: gap method with insulating dividers, double electrode method for cells, patch-clamp method for pieces
(patches) of membrane.
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 electrodes 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 voltage 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
FIGURE 16.20 Equivalent circuit
for the membrane of a small cell
with no spatial variation in its
electrical parameters.
Imembrane
G
C
V
418
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 constant over their entire surface whereas larger cells, such as neurons, can have potentials 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 distributedparameter 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 intracellular and extracellular 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 longitudinal resistance ri and r0, as well as inner and outer values for lonVmembrane
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 transversely across the insulation between the two co-axial conductors.
Although the mathematics of this model is complex, it is based on a
ELECTRIC CURRENT
AND
CELL MEMBRANES
Inside
riΔ(x)
Ii (x)
Im (x)
riΔ(x)
Ii (x + Δx)
Im (x + Δx)
VM (x,t)
Ii (x + 2Δx)
VM (x + Δ x,t)
Im (x + 2Δx)
gΔx
gΔx
gΔx
cΔx
cΔx
cΔx
V
V
roΔx
Io (x)
V
Io(x + Δx)
roΔx
x + Δx
FIGURE 16.21 A 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.
I o(x + 2Δx)
x + 2Δx
Outside
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
tM ⫽
cM
,
gM
(16.24)
and the space constant given by
l⫽
1
1(ri ⫹ r0)gM
.
(16.25)
Note how the units work out in Equation (16.24), with both cM and gM per-unit-length
constants so that their ratio has time units, whereas in Equation (16.25) the per-unitlength 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 membrane voltage difference VM from the resting potential decreases exponentially along
the axon in either direction according to
|x|
VM ⫽ VM
⫺
(0)e l
,
(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 calculate 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 (different 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 timedependence 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 locations at later times, but with a rapidly decreasing amplitude. If
E L E C T R I C A L P RO P E RT I E S
OF
N E U RO N S
FIGURE 16.22 The spatial variation
in the membrane voltage from
measurements along axons stimulated by a small current from an
electrode at x ⫽ 0.
419
t /τ = 1/16
normalized Vm
normalized Vm
1.2
1
0.8
0.6
0.4
0.2
0
–4
t/τ = 1
1
0.8
0.6
0.4
0.2
0
–2
0
x/λ
2
4
0
1
2
t/τ
3
4
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 16.24 Membrane voltage
changes during an action potential
(bold), together with sodium and
potassium ion conductances
across the membrane.
420
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 potentials, 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 resistor 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
VM not equal to 1).
The crux of the H-H model is the specification of the conductances GNa and GK.
Hodgkin and Huxley obtained these functions by fitting data from space-clamped measurements (eliminating the x-dependence, or the cable properties) that were also voltage-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 chemicals that block, or shut off, only one type of ion channel. From numerous measurements 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 associated 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 potential during which a second action potential 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
ELECTRIC CURRENT
AND
CELL MEMBRANES
Myelin
FIGURE 16.25 Myelin sheath
surrounding the axon with regularly
spaced nodes of Ranvier.
Axon
Nodes of Ranvier
different duration current pulses; and (5) accommodation, in which the membrane
adjusts to a sufficiently slow increase in current without producing an action
potential.
The H-H model was developed through studies of the giant squid axon, however,
most vertebrate neurons have a quite different structure that greatly modifies the nerve
conduction mechanism. All neurons are found in association with other nonneural
cells, known as supporting cells. In the central nervous system these are mostly glial
cells and in the peripheral nervous system they are predominantly Schwann cells. Most
of these supporting cells provide a myelin sheath that surrounds the axons of neurons
in segments, or internodes, that have narrow gaps, known as nodes of Ranvier, at regular intervals. The internodes are roughly 1 mm long, some 1000 times longer than the
nodes (Figure 16.25), and substantially change the electrical properties of nerve conduction. The myelin sheath provides a highly insulating layer effectively reducing
membrane currents, which are fairly well confined to the nodes where there is a much
higher density of channels than in the internodes. Myelin also greatly increases the
space constant ␭ so that the membrane potential changes occurring at one node spread
over many nearby nodes. Thus, a membrane potential depolarization occurring at one
node caused by local membrane currents will rapidly appear at nearby nodes triggering
membrane currents there as well. This type of signal propagation is known as saltatory
conduction (from the Latin for “to jump” and having nothing directly to do with salt)
because the membrane currents are triggered only at the nodes and not in a continuous
fashion along the axon. Action potentials generated by saltatory conduction travel at
much faster speeds (up to 100 m/s versus 20 m/s in squid giant axons) and myelinated
neurons also have much smaller diameters (20 ␮m versus 0.5 mm in squid giant
axons). A number of neuromuscular diseases, including multiple sclerosis (MS), affect
the myelin around axons.
6. MEMBRANE CHANNELS: PART II
In Part I of our discussion of membrane channels in the previous chapter, we
focused on the control and selectivity of voltage-gated ion channels. Now that we
have learned something about electrical circuits, we return to membrane channels
and discuss patch-clamp measurements of the electric currents through single channels. Patch-clamping was mentioned at the end of Section 4 above as a means of
space-clamping, or electrically isolating a patch of a cell membrane. Four types
of patch-clamps can be distinguished for use in recording the electrical activity of
single channels (Figure 16.26).
A micropipette tip pushed up against a cell membrane provides an initial low
resistance seal of about 50 M⍀. By applying suction, a gigaseal (G⍀ seal) is then
obtained where the patch is isolated from its surroundings by a huge resistance
value; this configuration is known as the cell-attached mode (A in Figure 16.26)
and is useful for studying voltage-gated channels or channels controlled by extracellular molecules supplied by the pipette. If the pipette tip is withdrawn pulling
on the membrane, the membrane will rupture at the pipette tip edge producing an
inside-out patch mode (B) if done in air or in the absence of divalent cations. By
then immersing the inside-out patch in an external solution, one can control the
ion content on the “cytoplasmic” side of the membrane.
M E M B R A N E C H A N N E L S : PA R T I I
421
FIGURE 16.26 Types of patch-clamps.
FIGURE 16.27 Single channel
recording from a patch-clamp.
Each step in current is the opening
and closing of a single ion channel.
I (pA)
0
1
Time (ms)
422
Alternatively, after forming a gigaseal, an additional pulse of suction or voltage
will open the membrane, exposing the cytosol to the pipette contents. This allows
whole-cell recording (C) to occur in which the pipette can introduce ions or chemicals or even proteins into the cell. Small cells can be studied quite well using this
method. For larger cells, pulling on the attached membrane, by surface tension, leads
to the formation of an outside-out patch (D), with the extracellular face of the membrane able to be immersed in an external solution.
Each of these methods results in a patch of membrane (possibly the entire small
cell) as a boundary between two controlled solutions, one within the pipette and one
external. At that point current flow across the membrane can be monitored by electrodes and amplifier circuitry. With only a few channels per square micron of membrane area, the sensitivity of the electronics is such that single channel recordings can
be made in which the flow of typically 5 picoamperes (1 pA ⫽ 10⫺12 A) of current
lasting typically 1 ms can be measured. Such a flow corresponds to about 30,000
monovalent ions through a single channel in the membrane. Our ability to measure
such small currents accurately hinged on the development of field-effect transistor
(FET) amplifiers which have very low noise characteristics.
Based on the macroscopic sodium channel currents measured for large membrane
surfaces discussed in the last section, one might guess that the single channel Na⫹ current recording would be just a miniature version of that continuous curve in time.
However, what is found in a single channel measurement is totally different. The current instead comes in individual discrete, rapid bursts of charge flow (Figure 16.27).
These pulses are spaced close together at times corresponding to a large macroscopic current and farther apart, on average, during smaller macroscopic currents.
Effectively, with a large number of identical channels, the current pulses add together
to give the continuous macroscopic current curve. An alternative way to consider this
ELECTRIC CURRENT
AND
CELL MEMBRANES
V (mV)
I (pA)
20
I (pA)
–100
100
Vm (mV)
Time (ms)
FIGURE 16.28 The upper voltage represents an applied depolarizing voltage
clamp, and the three single channel recordings, top three I versus t curves,
indicate a series of repeated measurements of current in synch with the
applied voltage pulse. When many repeated single channel recordings are
summed (bottom curve) the macroscopic aggregate current that results is
found to be the same as when measurements are made over a larger surface
area to give a macroscopic current signal directly in a single measurement.
–20
FIGURE 16.29 Current-voltage relations for
a single K⫹ channel. The conductance of
the channel can be obtained from the slope
of the dashed line.
is that repeated depolarization of the same channel, under identical conditions, will
elicit a seemingly random, different response each time; but when those responses are
summed, the average single channel current is found to mimic the macroscopic current
observed from a single simultaneous measurement on large numbers of such channels
(Figure 16.28).
The conductance of a single channel can be determined by measuring the channel current as a function of the membrane potential difference from the equilibrium
potential for that ion species (see Equation (16.23)). Figure 16.29 shows an example
of data from a K⫹ channel. The slope is the conductance for that channel under the
experimental conditions.
A simple model of ionic channels can be developed in which the channels exist
in only two possible states, closed (C) with zero conductance and open (O) with a
constant conductance. Figure 16.30 shows a hypothetical energy diagram for this
model. Note that the energy levels might well depend on the membrane voltage.
According to equilibrium thermodynamics, the ratio of the number of open to closed
channels is given by the Boltzmann factor as
N0
NC
⫽ e ⫺(E0 ⫺EC )/kB T.
(16.27)
The probability that a channel is open, P0, is given by the ratio
N0
P0 ⫽
N0
N 0 ⫹ NC
⫽
NC
N0
NC
⫹1
⫽
1
e ⫺(E0 ⫺EC)/kB T
.
⫽
⫺(E
⫺E
)/k
T
(E
e 0 C B ⫹ 1 1 ⫹ e 0 ⫺EC)ⲐkB T
(16.28)
A plot of P0 versus (E0 ⫺ EC)/kBT is shown in Figure 16.31. For large negative values of the abscissa, corresponding to a higher closed than open state energy, with the
difference large compared to thermal energies, all channels are open. In the opposite
limit of large positive values, all channels are closed. When the open and closed
energy values are equal, 50% of the channels are open. The energy levels of both
states change in response to the membrane potential.
In a further refinement of this model, various mechanistic models of the energies of
the two states can be assumed. For example, one simple scheme is to imagine a gating
molecule acting as a dipole and either spanning the channel, so that the channel is closed,
or not, so that the channel opens. Thus the rotation of a dipole—triggered by electrical
M E M B R A N E C H A N N E L S : PA R T I I
Energy
Eo
Ec
closed
open
FIGURE 16.30 Energy diagram for
two-state model of an ion channel.
423
1
0.8
Po
0.6
0.4
0.2
- -
0
–10
–5
0
(Eo-Ec)/kT
5
closed
+
+
open
10
FIGURE 16.32 A model for a
two-state dipole control of channel
opening and closing.
FIGURE 16.31 Probability of channel being
open as a function of the energy difference.
forces—controls the conductance of the channel (Figure 16.32). The interaction energy
can then be written in terms of the dipole and the membrane potential and an analysis
and comparison with data can lead to an estimate of the valence of the gating charge ze
on the dipole. Hodgkin and Huxley’s work showed that z is about 6 for the sodium channel, so that six positive charges are needed to shift from the cytosolic to the extracellular side of the membrane in order to give the observed voltage-dependence for the
gating. Equivalently 6 negative charges can shift across the membrane in the opposite
direction, or 12 charges could shift halfway across, and so on. Although some features
of the H-H model can be recovered from this simple model, multistate channel models,
with additional parameters, have also been developed.
CHAPTER SUMMARY
When placed in an electric field directed along a wire,
the free electrons in the conductor move randomly
about at thermal velocities (~106 m/s) while drifting
along the wire at very low speeds (~mm/s). The drift
velocity produces a net flow of charge Q, making up an
electric current I, defined as
I⫽
Requiv ⫽ R1 ⫹ R 2
(resistors in series)
¢Q
.
(16.4)
¢t
In a conductor, the current is proportional to the
applied potential difference V through Ohm’s law
V ⫽ IR,
(16.7)
where R is the electrical resistance of the conductor. As
current flows through a conductor, electrical energy is lost
thermal energy through joule heating at a rate given by
V2
P ⫽ IV ⫽ I 2R ⫽ .
R
(16.10)
Electrical circuits can be analyzed using two fundamental
rules: Kirchoff’s loop equation, stating that the net voltage
difference around any closed loop in a circuit is zero, and
424
the junction rule, stating that at any branch point in a circuit the total current into the branch point must equal the
total current flowing out. In simple circuits with either just
two resistors or two capacitors, we can develop the following rules for finding net R and net C values:
1
1
1
⫽
⫹ ,
Requiv R1 R2
1
1
1
⫹
⫽
C C1 C2
C ⫽ C1 ⫹ C2
( resistors in parallel) (16.14)
(capacitors in series)
(16.16)
(capacitors in parallel).
(16.18)
In circuits with series R and C elements, an analysis
finds that when discharging the capacitor the charge on
the capacitor and the current in the circuit both
decrease exponentially according to
Q ⫽ Q0 e
⫺
t
RC
,
ELECTRIC CURRENT
(16.20a)
AND
CELL MEMBRANES
I ⫽ I0 e
⫺
t
RC
,
(16.20b)
An analysis of the potential difference across a membrane with an imbalance in the concentration of ions on
both sides of the membrane leads to an equation, the
Nernst equation, for the potential difference across the
membrane due to the difference in ion concentration on
the inside (ci) versus outside (c0)
VK ⫽
c0
RT
log a b
ci
zF
(16.22)
where K is the example of potassium ions, R is the molar
gas constant, z is the ion valence and F is the Faraday
QUESTIONS
1. Some mistakes students make as they are learning
about circuits involve using wrong language to
describe situations, leading to or caused by conceptual misunderstandings. For example, it is fairly common to hear students say that current flows across a
resistor, or that voltage flows around the circuit, or to
ask what the voltage of the resistor is. What is wrong
with each of these statements?
2. Explain how it is that when you turn on an electric
light switch, the light comes on immediately even
though the electrons making up the electric current
travel at very slow speeds of only about mm/s.
Develop an analogy with water coming out of a full
hose when the valve is first opened.
3. If conductors cannot have electrostatic fields within
them, what is the mechanism that produces the force
on electrons within a conducting wire in a circuit
when there is an electric current flowing?
4. If free electrons in a conducting wire experience a net
force due to the electric field in the wire, why don’t
they accelerate continuously instead of traveling
along with a constant average velocity?
5. Explain the difference among resistivity, resistance,
conductivity, and conductance. Which are intrinsic
properties of a material and which depend on its size
and shape?
6. If a resistor of resistance R is connected to a battery
of voltage V, the equation for the power dissipated in
the resistor, P ⫽ I2R, implies that a larger resistor will
dissipate more energy and get hotter than a smaller
resistor. This is not true. Explain why not.
QU E S T I O N S / P RO B L E M S
constant. Using this basic idea, nerve impulses can be
modeled as combinations of such potential differences
that have different time behaviors in the Hodgkin–Huxley
and other models of nerve conduction.
To test these models, special techniques have been
developed to measure electrical properties of membranes
in order to relate them to their structure. In particular,
patch-clamping techniques allow scientists to study the
electrical properties of single special channels or pores,
made from individual proteins embedded in biological
membranes, by measuring pulses of current flow corresponding to the opening of a single one of these membrane channels. The time-varying voltage signal seen
from earlier voltage clamp measurements agrees with
both patch clamp measurements averaging over large
numbers of such pores, and with time-averaging single
current pulse measurements after repeated channel openings, as seen, for example, in Figure 16.28.
7. A homeowner keeps losing electric power during a
hot summer evening due to blown 20 A fuses. After
replacing the blown fuse several times and having the
same problem, he decides to use a 30 A fuse so it
won’t blow with the same electrical devices on. Why
is this a bad idea?
8. Copper wires covered with rubber-based insulation are
commonly used in household electrical wiring. These
wires come in different gauges, corresponding to different diameters of the copper wire, where increasing
gauge corresponds to decreasing diameter. Calculate the
resistance of 100 m length of 14 gauge wire (1.63 mm
diameter) and of 10 gauge wire (2.59 mm diameter).
According to National Electric Code standards, the
maximum current capacities of these two wires are 15 A
and 25 A. Which can carry more current?
9. Check that the SI units of the product of R and C are
seconds; verify that specific resistance times specific
capacitance also has units of seconds.
10. Two physics students are each measuring the RC
time constant of a simple series RC circuit. One of
them sets the initial voltage on the capacitor to 10 V
and measures the time for the voltage to drop to 5 V.
The second student, using the same circuit, sets the
initial voltage to 20 V and measures the time for the
voltage to drop to 10 V. Will these times be the
same? Why?
11. In circuit analysis, Kirchhoff’s loop equation is often
equated with conservation of energy, whereas
Kirchhoff’s branching equation for currents is often
equated with conservation of electric charge. Discuss
this statement.
425
12. Explain in words what an equivalent resistor means
when replacing some collection of resistors in a circuit by an equivalent resistor.
13. Explain why when using a multimeter as a voltmeter
its two wire leads can be simply put in parallel, or
across, the circuit element whose voltage is to be
measured, but when used as an ammeter this cannot
be done, but rather a wire leading to that circuit element must be “broken” so that the ammeter can be
inserted in series with it. Discuss this in words and in
terms of Kirchhoff’s equations.
14. A flashlight bulb acts as a small resistance when connected to a battery. If two identical bulbs are connected in parallel to the same battery will they be
brighter, dimmer, or the same brightness as when a
single bulb is connected to that same battery? Repeat
this when the two bulbs are placed in series across the
same battery.
15. In the previous question, does the battery supply
more, less, or the same current with two bulbs in parallel as when a single bulb is connected to the battery? Answer this when the two bulbs are placed in
series across the battery.
16. Given an unlimited supply of 100 ⍀ resistors, how
could you arrange a network of them to have an
equivalent resistance of 150 ⍀? Of 75 ⍀?
17. Discuss the meaning of the two parameters of the linear cable model of a neuron, the space and time constants. What do they tell us?
18. In the cable model, discuss in words the function of
each of the circuit elements.
19. What is the fundamental goal of a patch-clamp?
20. Can you think of any other physical processes like
membrane channel current that appear continuous on
one level, but are actually made up of discrete small
packets on a finer level?
Questions 21 and 22 refer to: Consider the circuit to the
right. The battery is a perfect source of emf. Treat A, B,
C, and D as bulbs of equal resistance.
B
A
C
D
21. Rank order the brightness of the bulbs in the circuit
shown.
22. Fill in the following table with “S” for same brightness, “D” for dimmer, “B” for brighter, and “O” for
goes out. “U” means that bulb is “unscrewed” for that
situation.
426
Situation
A
1
U
2
B
C
D
U
3
U
4
U
MULTIPLE CHOICE QUESTIONS
1. One end of a resistor in a simple circuit is at ⫹5 V; the
other is at ⫹3 V. Which one of the following is true? (a)
Electrons must be entering the resistor at the ⫹5 V end
and leaving at the ⫹3 V end. (b) Electrons leaving the
resistor have a higher kinetic energy than electrons entering the resistor. (c) Electrons at the ⫹3 V end have a
higher electric potential energy than at the ⫹5 V end. (d)
A current of 2 A must be flowing through the resistor.
2. A pure parallel combination of resistors has an equivalent (or effective) resistance of 2 ⍀. Which one of
the following is true? (a) The sum of the individual
resistances is 2 ⍀. (b) The sum of the reciprocals of
the individual resistances is 2 ⍀. (c) Each of the individual resistances is greater than 2 ⍀. (d) Each of the
individual resistances is smaller than 2 ⍀.
3. The statement, “The current in a resistor is directly
proportional to the potential difference across the
resistor,” is known as (a) Coulomb’s law, (b) Gauss’s
law, (c) Ohm’s law, (d) Ampere’s law.
4. The electrical resistance of a long piece of wire is R.
The wire is stretched to be twice as long and, because
the wire’s volume doesn’t change, its cross-sectional
area is halved. The electrical resistance of the stretched
wire is (a) R/2, (b) R, (c) 2R, (d) 4R.
5. A steady current flows through a resistor. An electron
in the current flow enters the resistor at the resistor’s
⫹5 V end and leaves at the resistor’s ⫹10 V end.
Which one of the following is true? (KE ⫽ kinetic
energy, PEE ⫽ electric potential energy.) (a) ⌬KE ⫽ 0,
⌬ PEE ⬍ 0, (b) ⌬KE ⬍ 0, ⌬ PEE ⫽ 0, (c) ⌬KE ⬎ 0,
⌬ PEE ⬍ 0, (d) ⌬KE ⫽ 0, ⌬ PEE ⬎ 0.
6. A voltmeter is used to read the potential difference
across the poles of a battery. The battery is rated at
20 V. The battery is connected in series to a switch,
an ammeter, and a resistor. When the switch is open,
the ammeter reads 0.0 A and the voltmeter reads
20.0 V. When the switch is closed the ammeter reads
1.0 A and the voltmeter reads 19.0 V. Which one of
the following is most likely to be the explanation for
this result? (a) The ammeter has too little resistance.
(b) The voltmeter has too much resistance. (c) The
battery has an internal resistance of 1.0 ⍀. (d) The
resistor in the circuit has a resistance of 20.0 ⍀.
7. A battery has an emf of 10 V and an internal resistance of 1 ⍀. When the battery is connected to a combination of resistors, a perfect ammeter reads a
current of 1 A leaving the positive pole of the battery.
ELECTRIC CURRENT
AND
CELL MEMBRANES
At the same time, a perfect voltmeter placed across
the poles of the battery will read a potential difference
of (a) 11 V, (b) 10 V, (c) 9 V, (d) 1 V.
2
1
3
8. Two resistors are in parallel as shown in the figure
above. When the current at point 1 is 0.15 A and the
current at point 2 is 0.05 A, what is the current at
point 3? (a) 0.15 A, (b) 0.10 A, (c) 0.05 A, (d)
between 0.05 A and 0.10 A.
9. Two resistors are connected to an ideal battery in
series. Resistor 1 has a potential difference across it of
10 V and resistor 2 has a potential difference across it
of 20 V. Now, the two resistors are connected to the
same battery in parallel. The potential difference
across resistor 1 (a) is now 10 V, (b) is now 20 V,
(c) is now 30 V, (d) cannot be calculated because the
resistances aren’t given.
12. V1 must read (a) 0 V, (b) 5 V, (c) 10 V, (d) some value
that depends on the actual emf of the battery and the
actual resistances.
13. V3 must read (a) 0 V, (b) 5 V, (c) 10 V, (d) some value
that depends on the actual emf of the battery and the
actual resistances.
14. A1 must read (a) 1 A, (b) 2 A, (c) 3 A, (d) some value
that depends on the actual emf of the battery and the
actual resistances.
15. A2 must read (a) 1 A, (b) 2 A, (c) 3 A, (d) some value
that depends on the actual emf of the battery and the
actual resistances.
16. Two light bulbs, one rated at 50 W and a second rated
at 100 W, are both supposed to be connected to a 110 V
source of emf. Which one of the following is true? The
50 W bulb has (a) twice the resistance as the 100 W
bulb, (b) four times the resistance of the 100 W bulb,
(c) half as much resistance as the 100 W bulb, (d) one
quarter as much resistance as the 100 W bulb.
17. A bulb (i.e., a resistor) is connected in series to a switch,
a battery, and an uncharged capacitor. At t ⫽ 0, the
switch is closed. Which of the following best describes
the brightness of the bulb as a function of time?
Brightness
Brightness
Time
(a)
Brightness
Questions 10 and 11 refer to the figure above
10. What is the equivalent resistance of the circuit shown
in the figure to the right? Each resistor is 1 ⍀. (a) 4
⍀, (b) 1.67 ⍀, (c) 0.60 ⍀, (d) 0.25 ⍀.
11. In the circuit shown if the battery supplies 3 V and
each resistor is 1 ⍀, what is the current through the
resistor in the middle branch? (a) 3 A, (b) 2.4 A,
(c) 1.2 A, (d) 0.6 A.
QU E S T I O N S / P RO B L E M S
Brightness
Time
(c)
Time
(d)
18. Which of the following best describes the potential
difference across a capacitor that is connected in series
to a resistor and a source of emf that is sequentially
⫹V for time T, then 0 for time T, and so on, when T is
small compared with RC? Vertical axes are potential
difference, horizontal axes are time. The lighter plots
are the emf, the bolder plots are the capacitor voltage.
(a)
Questions 12–15 refer to the figure above. The A’s are
perfect ammeters, the V’s are perfect voltmeters, the battery is a perfect source of emf, and the resistors are equal.
V2 reads 5 V and A3 reads 1 A.
Time
(b)
(c)
(b)
(d)
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PROBLEMS
1. What is the equivalent current in a solution of
monovalent ions flowing through a capillary
tube such that 1 mM of ions leaves the tube each
second.
2. A capacitor with a charge of 5 ␮C has its terminals
shorted by a metal wire so that the charge flows off
within 2 ␮s. What is the average current flowing during that time?
3. What is the average current when all the sodium channels on a 100 ␮m2 patch of muscle membrane open
together for 1 ms? Assume a density of 50 sodium
channels per ␮m2 of surface and a flow rate of 1000
ions per ms through each channel.
4. Calculate the conductance and the resistance of a
10 m length of 14 gauge copper wire, which has a
diameter of 1.63 mm. If this wire is connected
directly to the terminals of a 12 V dc power supply,
shorting it, how much current will flow assuming
the power supply can deliver an unlimited amount
of current?
5. A 1 cm3 cube of gold (␳ ⫽ 1.61 ⫻ 10⫺8 ⍀m) is
drawn out into a uniform cylinder of 20 m length.
What is its electrical resistance?
6. Two 100 m 14 gauge wires (1.63 mm diameters), one
of copper and one of aluminum, are soldered together
and the 200 m wire is then connected to a 6 V dc
power supply with unlimited current.
(a) How much current flows in the wire?
(b) What is the potential across each 100 m section
of wire?
(c) How much power is developed in each section of
wire?
7. A 1000 W heater runs from a 100 V dc power
supply.
(a) How much current flows in its heating cable
wire?
(b) What is the resistance of the wire?
8. An electric eel, found in the rivers of Brazil, can
discharge lethal currents of 1 A at 400 V. How much
power does the eel generate?
9. An immersible heater coil is to be designed to heat an
insulated container with 4 liters of distilled water
from 20° to 50°C in less than 30 min.
(a) How much energy must be input to heat the
water to this temperature?
(b) To heat the water, what minimum power must be
supplied?
(c) If a 12 V power supply is to be used, what
minimum current must flow in the heating coil?
(d) What must be the total resistance of the heating
coil? Is this a maximum or minimum resistance
to heat the water in 30 min or less?
10. Determine the equivalent resistance between points A
and B in the following circuit.
428
5 kΩ
5 kΩ
A
4 kΩ
1 kΩ
B
8 kΩ
3 kΩ
11. Given the network of equal 1 k⍀ resistors shown
below, compute its equivalent resistance and the current drawn from the 12 V power supply. (Hint:
Combine resistors in stages using the simple rules for
series and parallel combinations of resistors.)
12. Analyze the circuit shown below to find the currents
flowing through and the power generated in each
resistor.
1.5 kΩ
6V
4 kΩ
1.5 kΩ
2.5 kΩ
13. A single 1 M⍀ resistor is connected across a power
supply. An ammeter is inserted to measure the current
out of the battery. If a voltmeter with 10 M⍀ resistance is used to measure the voltage across the
resistor, what will be the percent change in the current
reading on the ammeter when the voltmeter is connected across the resistor?
14. Find the reading that each of the (ideal) meters would
have in the following circuit.
10 V
A2
3 kΩ
V1
1 kΩ
V2
10 kΩ
A1
1 kΩ
5 kΩ
ELECTRIC CURRENT
AND
CELL MEMBRANES
15. Analyze the following circuit to find the current flowing through the 10 k⍀ resistor.
12 V
5 kΩ
10 kΩ
2 kΩ
6V
16. Find the current in the central branch of the following
circuit.
12 V
3 kΩ
5 kΩ
10 kΩ
12 V
17. RC time constants can be easily estimated by measuring
the time (known as the half-time) for the capacitor voltage to decrease to half of some arbitrary starting value
when discharging through a resistor. From Equation
(16.12a), the voltage across the capacitor will vary as
t
V(t) ⫽ V0 e
⫺
RC
.
Show how a single measurement of the half-time can
be used to determine the RC time constant. (Hint:
Substitute V(t) ⫽ V0/2.)
18. A 100 ␮F capacitor wired in a simple series RC circuit is initially charged to 10 ␮C and then discharged
through a 10 k⍀ resistor.
(a) What is the time constant of the circuit?
(b) What is the initial current that flows?
(c) How much charge is left on the capacitor after
1 time constant?
(d) What is the current after 1 time constant?
(e) How much charge is left on the capacitor after
3 time constants have elapsed and what current is
flowing then?
19. A simple RC series circuit has a 100 ␮F capacitor.
(a) If the time constant is 50 s, what is the value of the
resistor?
(b) Suppose that a second identical resistor is inserted
in series with the first. What is the new time constant of the circuit?
QU E S T I O N S / P RO B L E M S
(c) Suppose the second identical resistor is placed in
parallel with the first resistor, still connected to
the capacitor. What is the new time constant in
this case?
20. Consider a defibrillator, acting as a 32 ␮F capacitor
and a 47 k⍀ resistor in a series RC circuit. The circuitry in this system applies 5000 V to the RC circuit
to charge it.
(a) What is the time constant of this circuit?
(b) What is the maximum charge on the capacitor?
(c) What is the maximum current in the circuit during
the charging process?
(d) What are the charge and current as functions of
time?
(e) How much energy is stored in the capacitor when
it is fully charged?
21. We’ve seen that the Earth’s atmosphere is able to act
as a capacitor, with the ground and the clouds acting
as plates with an air gap in between. Under certain
circumstances air can be made to conduct, so that
electric charge can flow from the clouds to the ground
in what we call a lightning bolt. Assuming that the
clouds are distributed around the entire Earth at a
fixed distance of 5000 m above the ground of area
4␲R2Earth, where REarth ⫽ 6400 km, the resistance of
the air between the clouds and the ground is calculated to be R ⫽ 300 ⍀.
(a) Assume that the charge is distributed spherically,
so that V ⫽ k(Q/r) and therefore ⌬V is the difference in potential between the lower plate (the
Earth’s surface) and the upper plate (the clouds).
In addition, assume that in a typical day, 5 ⫻ 105
C of charge is spread over the surface of the Earth.
What is the potential difference between the
clouds and the ground?
(b) What is the capacitance of the Earth–cloud
capacitor?
(c) If the charge on the clouds is discharged through
the air, what is the capacitive time constant for this
discharge?
(d) How many lightning strikes does this amount of
charge correspond to if each lightning strike contains about 25 C of charge?
(e) Approximately how long would it take the
Earth–cloud capacitor to discharge to 0.1% of its
initial charge?
(f) Assuming that the charge is immediately replenished as soon as the discharge process ends,
approximately how many lightning bolts are there
per day?
22. Fill in all the steps in the calculation of the number of
ions crossing a membrane channel when it opens (see
Section 16.3 following Example 16.5). Now, using
those same numbers, calculate the total number of
moles of charge crossing the membrane when the
429
membrane of a spherical cell with a radius of 10 ␮m
completely depolarizes. If this charge were all K⫹
leaving the cell, calculate the fraction of the K⫹ present in the cell interior that crosses the membrane
when it depolarizes. (Hint: You need to calculate the
number of moles of K⫹ inside the spherical cell and
the number on the surface of the membrane, all of
which is assumed to cross the membrane when it
depolarizes; see Table 16.2.)
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23. Check the values of the Nernst potential in Table 16.2
using Equation (16.22).
24. Show that Equation (16.28) leads to the plot shown in
Figure 16.31. In particular, show that the P0 values
for E0 ⫽ EC and for large and small values of the difference (E0 ⫺ EC) are correct.
ELECTRIC CURRENT
AND
CELL MEMBRANES