Electric Current and Cell Membranes - if.ufrj.brcoelho/Newman/Newman16.pdfpotential difference...

E L E C T R I C C U R R E N T A N D R E S I S TA N C E 401

J. Newman, Physics of the Life Sciences, DOI: 10.1007/978-0-387-77259-2_16, © Springer Science+Business Media, LLC 2008

Thus far in our study of electricity, we have essentially confined our attention toelectrostatics, or the study of stationary charges. Here and in the next three chapterswe show some of the new phenomena that arise when charges move. We begin thischapter by generalizing our discussion to allow the flow of electric charges, knownas an electric current, and we give a semiempirical derivation of Ohm’s law.Electrical measurement methods and devices are described as an application ofOhm’s law. More realistic models for a capacitor are then developed in a continuedstudy of cell membranes in which electric charge can passively leak across the mem-brane. We give an overview of nerve structure and functioning and the spatial andtemporal properties of the neuron membrane potential are detailed for both the qui-escent and active states. The chapter concludes with a discussion of the electricalproperties 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, wehave not discussed the consequences of channels on the electrical properties of mem-branes. Membranes act as capacitors, storing charge and electric potential energy, butbecause of “leakage” of charge through channels, membranes are not the ideal capac-itors treated in the last chapter. In order to discuss more realistic models for mem-brane electrical properties we first need to introduce some concepts related to theflow of electric charge.

Figure 16.1 shows a conducting wire attached at time zero between the plates ofa previously charged air-spaced capacitor. Before the wire is connected we havealready seen that there is an electric field between the plates of the capacitor, but nocharge flows because the air is a good electrical insulator. As soon as the wire is con-nected, there will be an electric field in the wire that will drive the free electronstoward the positive capacitor plate, discharging the capacitor. The electric current inthe wire is defined as the time rate of flow of charge along the wire

(16.1)

where the direction of the current is chosen by convention as opposite to the flow ofthe electrons. Thus, the electric current flows from the positive to negative plates ofthe capacitor in our example. The SI unit for electric current is the ampere (A), givenby 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 thewire is not constant in this situation because as the charge drains off the capacitor

I �¢Q

¢t,

16Electric Current and CellMembranes

plates, the electric field that drives the electric charges decreases. If the initial chargeon 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, decreasingcontinuously as the charge is drained from the capacitor plates. We show below howto find the actual time dependent current flowing in this simple electric circuit.

Unlike the electric fields of previous chapters, the electric field driving thecharges through the wire is not an electrostatic field. In fact, as we have seen, elec-trostatic fields cannot exist within a conductor. The electric field that drives the elec-tric current, on the other hand, does exist within the conductor and is responsible forpushing the charge making up the current. This example illustrates that without asource of energy to maintain net charge on the plates of the capacitor, both the elec-tric field in the wire and the current flow rapidly decrease to zero.

After charging the capacitor in Figure 16.1, we can think of the discharging of thecapacitor as the conversion of electric potential energy to the kinetic energy of the elec-trons in the wire connecting the plates. As we show at the end of this section, the kineticenergy of the free electrons making up the current is then converted into heat via colli-sions with the metal atoms of the wire. The discharging of the capacitor occurs rapidlyand therefore there is only a pulse of electric current in this case. In order to maintain aflow of electric charge, an external source of energy per unit charge, traditionally calledan 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 auniform wire connected between its terminals. If the battery were simply a capacitor asin Figure 16.1, the initially separated positive and negative charges would quickly can-cel each other out as charge flows along the wire and there would be no further change.Batteries convert chemical energy to electrical energy to continually maintain a separa-tion of charge and supply a fixed voltage between their terminals. This is shown on theright side of the figure where the varying voltage is shown as it might be measuredaround the circuit, with the battery increasing the voltage each time around. A very goodanalogy is the flow of water due to gravity where the potential energy decreases as thewater flows down hill and can only be restored by a pump of some kind, playing the roleof the battery, to increase the height and thus the potential energy of the water. In ourcase 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 isproduced by a uniform electric field in the wire maintained by the battery and given byE � 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 amicroscopic picture of a collection of free electrons in the conducting wire and theforces acting on them. In the absence of an external electric field, the thermal energyof the free electrons causes them to diffuse about in a random walk traveling at veryhigh speeds of about 106 m/s and making random collisions with the atoms of themetal wire (see the discussion in Chapter 2). The average velocity of the electrons, asopposed 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 itshigh-speed random walk motion, a free electron will experience an acceleration (in

the direction opposite to the electric field because of thenegative electric charge) given by

(16.2)

where e and m are the charge and mass of the electron. Thisacceleration lasts until the electron makes a collision with ametal atom causing it to veer off in another random directionat high speed, accelerating again according to Equation (16.2).

a �Fm

�eEm

,

402 E L E C T R I C C U R R E N T A N D C E L L M E M B R A N E S

E

I

+Q–Q

FIGURE 16.1 Two chargedconducting plates connected by a conducting wire at time zero.

+ -

voltage

Distance around circuitfrom battery

wire

battery

wire

battery

FIGURE 16.2 (left) A battery withits terminals connected by auniform wire. (right) Voltage as afunction of distance around thecircuit showing the decreasingvoltage in the wires and the boostin voltage across the battery fromchemical energy every time aroundthe circuit loop.

The mean time between collisions, �, is so short that the electrons only acquire a very slowdrift velocity of about 10�3 m/s given by

(16.3)

If the number of free electrons per unit volume, or number density, in the wire is n andthe wire has a cross-sectional area A, then the net free charge in a short length of the wirel is �Q � nAle (Figure 16.3). To find the current in the wire, we must divide �Q by thetime required for all of that charge to move a distance l down the wire, �t � l/vdrift, to find

(16.4)

Substituting from Equation (16.3), the electric current is

(16.5)

Defining the conductivity � of the wire, an intrinsic property of the material, to be

we can rewrite Equation (16.5) as

(16.6)

where G is known as the conductance.Solving for V, this can be rewritten in terms of the resistance R

(16.7)

where

The resistivity of the material is given by the inverse of the conductivity,

both intrinsic parameters. This definition is made in analogy with the equalitybetween the resistance and the inverse of the conductance

except that both of these quantities are dependent on the size and shape of the mate-rial, so that they are extrinsic parameters, unlike the intrinsic parameters dependingonly 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 thepotential difference applied between the ends of the wire. This linearity of current withapplied voltage (Equation (16.7)) is known as Ohm’s law. A plot of the current througha wire as a function of the voltage across the wire is shown in curve A of Figure 16.4.The linear plot is characteristic of an ohmic (or linear) circuit element. Another equiv-alent statement of Ohm’s law is that the resistivity of the material remains a constant,independent of the applied voltage.

R �1

G,

r�1s

,

R �1

G�

1s

L

A�r

L

A.

V � IR,

I � sA

LV � GV,

s�ne2t

m,

I �ne2tA

mLV.

I �¢Q

¢t�

nAle

(l/ vdrift )� nAevdrift.

vdrift � at �eEm

t �eVt

mL.


l

Avd

FIGURE 16.3 Free charge in a wireof cross-sectional area A andlength l traveling with a driftvelocity vd.

I

V

Slope = 1/RA B

FIGURE 16.4 The I–V curve for an ohmic circuit element (A) and asemiconductor diode (B).

The SI unit for resistance is the ohm (), where 1 V/A �1 (read as 1 ohm). Units for resistivity are then given as -m and for conductivity as (-m)�1. The unit for conduc-tance, the reciprocal of resistance, is the �1 which is alsoknown as the siemens (S). Table 16.1 lists some values forresistivity of various materials. A wire made from a metal willhave a very low resistance value. For example, a 1 m length of1 mm diameter copper wire has a resistance of only 0.02 .Simple devices known as resistors (shown in Figure 16.5) aremanufactured to have various resistance values. The symbol

is used to represent a resistor in a schematic or circuitdiagram such as the one shown in Figure 16.6. Connectingwires have negligible resistance, so that their length and shapeare usually not important in a circuit diagram or in the actualcircuit itself.


Table 16.1 Resistivities of Various Materials (20°C)

Material Resistivity, ( .. m)

Conductors

Aluminum 2.8 � 10�8

Copper 1.7 � 10�8

Iron 10. � 10�8

Mercury 96. � 10�8

Silver 1.6 � 10�8

Tungsten 5.6 � 10�8

Ionic materials

Water (distilled) ~2 � 105

Fresh water ~5 � 102

Sea water ~0.3

Cytoplasm ~0.5

Fatty tissue ~15

Semiconductors

Germanium ~0.5

Silicon ~2. � 103

Insulators

Air (dry) 4 � 1013

Glass 1010 � 1014

Rubber 1013 � 1016

Example 16.1 How much electric current flows through water contained in aninsulating tube 10 cm long and 5 cm in diameter when a 100 V potential differ-ence is applied across the ends of the tube using electrodes inserted at eitherend? Ignore any complications from the metal electrode–water contact and dothe 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, whereR is the resistance between the two electrodes supplying the 100 V potentialdifference. Using the relation between resistivity and resistance, and the dimen-sions of the water tube, we find that

FIGURE 16.5 An assortment of resistors.

R1

R2

FIGURE 16.6 A simple circuitdiagram showing a batteryconnected to two resistors, onewired after the other.

Ohm’s law is not a fundamental law on par, for example, with Newton’s laws. Itis a heuristically derived statement that the current and voltage are proportional in aconductor. Many electrical components, such as diodes, transistors, operationalamplifiers, 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 bothresistors and nonlinear circuit elements in them.

Next we briefly consider the general topic of electrical energy and power. In thesimple circuit of Figure 16.2, the battery terminals are maintained at a constantpotential difference by chemical energy with the positive terminal at potential Vbatterywith respect to the negative terminal at V � 0. When the wire of length L is con-nected between the terminals, an electric current flows from the positive to negativeterminal. If we plot the electric potential as a function of position along the wire(Figure 16.7), we see that it decreases linearly from the battery voltage at the posi-tive terminal to zero at the negative terminal of the battery. A (positive) charge �Qflowing from the positive to negative battery terminal flows down this potential hillso that the decrease in electric potential energy is

(16.8)

Because in a time �t, the charge flowing in the wire is �Q � I �t, the rate at whichelectric energy is lost is given by the electric power P,

(16.9)

The SI unit for electric power is the Watt, just as for all other powers, as can be ver-ified by substituting units for IV, 1 A � 1 V � 1 CV/s � 1 J/s � 1 W.

If we examine the flow of energy in this example, stored chemical energy of thebattery is used to maintain a constant potential difference between the battery termi-nals. This constant V produces a constant E field within the wire that, in turn, main-tains a constant drift velocity for the charges. Thus, the kinetic energy of the chargesremains constant along the wire, although energy is continually lost through colli-sions. As charge flows along the wire and down the potential hill of Figure 16.7, thepotential energy loss at a rate P appears as thermal energy of the wire causing a tem-perature increase. This transfer of energy occurs through the collisions with the arrayof metal atoms in the wire while the drift velocity is maintained by the constant elec-tric field using energy supplied by the battery. The electrical energy is said to be lostbecause the entire process is irreversible. As we have seen in our study of thermody-namics, a loss of potential energy of any kind to heat cannot be a truly reversibleprocess.

Other expressions can be obtained for the power in terms of the resistance of thewire in our example. Using Ohm’s law, Equation (16.7), to eliminate either V or I, weobtain

(16.10)P � IV � I 2 R �V 2

R.

P �¢PEE

¢t�

¢Q

¢tV � IV.

¢PEE � ¢Q V.


Corresponding values are then, for distilled water, R � 0.1 � 1012 and I �1.0 nA; for fresh water, R � 2.5 � 108 and I � 0.4 �A; and for sea water, R� 0.15 M and I � 0.67 mA. The huge increase in current of almost a factor ofone million is due to the increase in ion content of the sea water versus freshwater versus distilled water.

R � rL/A � r 0.1

p(0.05/2)2� 51r.

Vbattery

V

distance along wire L

FIGURE 16.7 The voltage,measured with respect to thenegative terminal of the battery,along the uniform wire of length Lin Figure 16.2.

This conversion of electrical energy to thermal energy in a resistor is known as Jouleheating. 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 heatingcan 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 Example16.1. Also, find the rate at which the water temperature increases if no heat is lostto the surroundings.

Solution: The power calculation is straightforward using, for example, V2/R, tofind powers of 0.1 �W (distilled water), 40 �W (fresh water), and 67 mW (seawater). If none of the input power is lost, it is all converted to heat in the water.The water temperature will rise at a rate determined from

where P is the I2R Joule heating. The volume of water is given by �r2L �(3.14)(.025)2(0.1) � 2.0 � 10�4 m3, so that the mass of the water is about 0.2 kg,roughly independent of the salt concentration. Using a specific heat of4180 J/(kg°C), we find rates of temperature increase of 1.2 � 10�10°C/s (for dis-tilled water), 4.8 � 10�8°C/s (for fresh water), and 8.0 � 10�5°C/s (for sea water).These heating rates are quite negligible, taking several hours to heat the sea water1°C. However, if the tube length is decreased by a factor of 10 and the tube diam-eter is increased by a factor of 10, then the resistance will decrease by a factor of1000, and both the current and power will increase by that same factor. In this casethe heating is appreciable, increasing the sea water temperature by about 5°C/min.

¢Q

¢t� mc

¢T

¢t� P,

2. OHM’S LAW APPLICATIONS AND ELECTRICALMEASUREMENTS

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 analyzesome basic circuits. There are three common types of electric meters, often packagedin a multipurpose device known as a multimeter. By flipping a switch this device canmeasure current (as an ammeter), voltage (as a voltmeter), or resistance (as an ohm-meter). Although today these devices consist of complex semiconductor components,the fundamental principles of the devices can be more simply explained. Given a sim-ple circuit consisting of a battery and resistor as shown in Figure 16.8, how can oneuse a multimeter to measure the current in the circuit, the voltages across the batteryor resistor, and the resistance value of the resistor?

VoltmeterV

AAmmeter

R

OhmmeterR

R

a b c

A B

FIGURE 16.8 Measurement of (a) the current through R, with an ammeter inserted intothe 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 fromthe circuit, as shown above.

Any electrical measuring device has its own internal resistance that must bedesigned to minimize the impact of the presence of the meter on the electrical prop-erties being measured. To measure the current in the circuit of Figure 16.8a, the mul-timeter must be set to act as an ammeter and be inserted into the circuit by“breaking” a wire (actually by replacing the one wire between the resistor and bat-tery with two wires) and inserting the meter “in series” with the resistor. Being “inseries” means that the same current must flow through the ammeter as flows throughthe resistor; there is no other path for the current to follow. However, the presenceof the ammeter, with its internal resistance, affects the total resistance in the circuitand thereby the current. We would like to “analyze” this circuit; that is, we wouldlike to write the equations that allow us to predict the current the ammeter wouldmeasure 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 inthe circuit diagram, one mentally “travels” around a closed loop, adding and sub-tracting the potential increases and decreases algebraically as the loop is traversed.The sum must add to zero because on returning to the starting position, the potentialhas that same starting value and thus the potential difference around any closed loopmust be zero. In using the loop method, care is needed in choosing the proper alge-braic sign for the potential difference across each circuit element. For batteries thepotential increases when going from the � to � terminal across the device, whereasfor resistors, the potential drops in going across the resistor in the direction of the cur-rent flow according to Ohm’s law. Whichever direction one chooses to go mentallyaround a loop, a consistent set of potential differences must be summed to zero forthe 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 andsubtracting the appropriate voltages using Kirchhoff’s loop equation for circuit (a) inthe figure to obtain

or

where

(16.11)

In this equation, V is positive because we are “traveling” from the � to � terminal,and the IR voltages across resistors are both decreases (drops), taken as negative,because we are “traveling” around in the direction of the actual current flow from the� terminal of the battery. Our answer for this circuit is actually an example of a gen-eral result when any two (or more) resistors are connected in series:

The equivalent resistance of resistors in series is the sum of their individualresistances.

It also suggests that for an ammeter to have a negligible effect on the current in theoriginal circuit, it must have a very small resistance, certainly negligible compared tothe resistance in the circuit. Modern ammeters have a very low resistance, typicallyless than 1 . Given values for V and R, the equation above predicts the measuredammeter current.

In order to measure the voltage across any component in a circuit, a multimeteris set to act as a voltmeter and needs to have its terminals connected across that cir-cuit element as shown in Figure 16.8b to measure the voltage across the resistor. Thevoltmeter resistance is said to be “in parallel” with resistor R because both elementshave the same potential difference across them. However, the current flowing out of

R equiv � R � R ammeter. (resistors in series)

V � IRequiv,

V � IR � IRammeter � 0,

O H M ’ S L AW A P P L I C AT I O N S A N D E L E C T R I C A L M E A S U R E M E N T S 407

the positive terminal of the battery, when arriving at pointA in the figure, divides with part of the current flowingthrough each “branch” of the circuit later to recombine atpoint B. This is our first example of a multiloop circuit,one in which the same current does not flow through allthe circuit elements, and we digress further to show howit can be analyzed.

Consider the circuit shown on the left in Figure 16.9,similar to that of Figure 16.8b because the voltmeter isrepresented by a resistor in parallel with the original resis-tor. Using Kirchhoff’s loop equation to analyze this

circuit, we can write down several equations depending on the chosen loop:

clockwise around the outer loop, starting from B;

clockwise around the lower loop, from B; (16.12)

clockwise around the upper loop, from B.

Clearly these equations are not all independent, because, for example, subtracting thesecond from the first results in the third. An additional independent equation can beobtained by noting that at points A and B (branch points) where the current divides,by conservation of electric charge we must have that

(16.13)

where I is the current from the battery (see Figure 16.9 left). This is an example of asecond more general rule, known as Kirchoff’s junction rule, which states that at abranch point (or junction) where several wires come together, the total current enter-ing the branch point must equal the total current leaving that point. Clearly this is aconsequence of the general law of conservation of electric charge. Solving the firsttwo equations of Equations (16.12) for each of the currents I1 and I2 and substitutinginto Equation (16.13), we can write that

where Requiv is the single equivalent resistor that, when connected across the samebattery voltage V will cause the same current I to flow from the battery (see the rightside of Figure 16.9), so that V � I Requiv. Dividing by V, we obtain

(16.14)

showing the general rule for resistors in parallel:

The inverse of the equivalent resistance of resistors in parallel is the sumof the inverses of individual resistances.

Returning to the measurement of the voltage across a resistor in the circuit ofFigure 16.8b, by putting the voltmeter in parallel with the resistor the equivalentresistance seen by the battery will change (actually, it will always decrease; can youshow this from Equation (16.14)?) and therefore so will the current flowing out of thebattery (it will always increase in such a circuit). The excess current will be drawninto the voltmeter loop of the circuit. The battery current is entirely determined bythe “load”, or equivalent resistance, on the battery from V � IRequiv. To avoid chang-ing the battery current significantly, the voltmeter must have a very high resistance,so that it draws negligible current and the equivalent resistance is essentially that ofthe circuit, R. Modern voltmeters have resistances of about 10 M (1 M � 106 ).

1

Requiv�

1

R1�

1

R2, (resistors in parallel)

I �V

Requiv�

V

R1�

V

R2,

I � I1 + I2,

I2R2 � I1 R1 � 0,

V � I2 R2 � 0,

V � I1 R1 � 0,


R1

R2

V

I1

I2

V

Requiv

I I

BA

FIGURE 16.9 A simple circuit withtwo resistors in parallel.

A multimeter can also function as an ohmmeter when directly connected to bothsides of (across) a resistor which has been removed from the circuit, as in Figure16.8c. By using an internal battery to send a known current through the resistor andby measuring the voltage across the resistor, the ohmmeter directly measures itsresistance.


Example 16.3 Find the current that flows through each of the resistors shown inthe circuit of Figure 16.10. Also determine the power generated in each resistor.

Solution: In solving circuit analysis prob-lems it is important to first take a carefullook at the “lay of the land” or the circuit’sbasic “topology.” In this example, the 12 Vbattery is the only source of current in thecircuit and so it sends current out of its �terminal that then divides at the lower leftbranch point, some traveling through the5 k resistor and the rest traveling throughthe 1.5 k and 2.5 k resistors, which arein series with each other. The currents inthese two branches (the 5 k branch andthe (1.5 k � 2.5 k) � 4 k branch) recombine in the upper right corner andtheir sum, the net battery current, then travels through the 2 k resistor and returnsto the � terminal of the battery. It is very important for you to be able to understandand eventually generate this type of qualitative analysis before going to equationsin order to find values for the currents.

With the understanding of the previous paragraph, we can solve this problemin a simple straightforward manner, by finding the total equivalent resistance inthe circuit from the following: (1) first, the 1.5 k and 2.5 k are in series andtogether have a net resistance of 4 k shown on the left below; (2) then the 4 kand 5 k are in parallel with each other (do you see why?), so that their equiva-lent resistance R is given by 1/R � 1/4k � 1/5k, giving R � 2.22 k, shownin the middle below; (3) then the 2.22 k and the 2 k are in series with eachother yielding a net resistance in the circuit of 4.22 k, shown on the right.

The circuit on the right tells us that the current out of the battery is just I �(12 V/4.22 k) � 2.84 � 10�3 A � 2.84 mA. All of this current passes throughthe 2 k resistor because it is in series with the battery, but each of the other resis-tors only gets part of this current. To find how the current divides, we can workbackwards in the set of figures just above. The current divides at the branch pointso that the voltages across the 5 k and equivalent 4k resistor (see the left figureabove) are equal because the two branch points have a fixed potential V betweenthem whether we “travel” through the 5 k or 4 k resistor. This implies that

V � (I5 k 5 kÆ) � (I4 k 4 kÆ)

12 V 2 KΩ

5 KΩ

1.5 KΩ 2.5 KΩ

12 V 2 kΩ

5 kΩ

4 kΩ

12 V 2 kΩ

2.22 kΩ

12 V 4.22 kΩ

FIGURE 16.10 Circuit for Example16.3. Which resistors are in series orparallel with the others?

(Continued)

The preceding example was solved by simply using the rules for combiningvarious resistors in series and parallel. There are more complex circuits where thistype of analysis is not possible and Kirchoff’s loop equation must be used. The nextexample has such a circuit.


so that I5k/I4k � 4/5. But we know that the total current, I5k � I4k, is 2.84 mA,so that we can find the individual currents from either the previous two equationswith their two unknown currents, or from the following simple argument. Bydividing 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 kand (5/9) of the total current, or 1.58 mA, flows through the 4 k equivalentresistor. Finally returning to the original circuit, each of the 1.5 k and 2.5 kresistors have I4k � 1.58 mA flowing through them. You should check that theseresults are consistent and add up properly; follow each current around the orig-inal circuit and check Kirchoff’s junction rule.

We finish this problem by noting that the power generated in each resistor isgiven by P � I2R, so that if we know the values of the currents and resistors wecan 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 theresistors. Check this yourself !

Example 16.4 Find the current flowing though each resistor of the following circuit.

Solution: In this case, because of the sec-ond battery in the circuit we cannot simplycombine resistors in series and parallel butmust use Kirchoff’s loop equation. Usingthe set of labeled currents, which can bechosen arbitrarily as long as they areconsistent, we can write down two loopequations to allow us to solve for the twounknown currents labeled I1 and I2 in thefigure. We have already implicitly used thejunction equation in choosing the sum ofthe two currents from the batteries as thecurrent in the central branch of the circuit. Follow the currents to the right junctionpoint and check that they are self-consistent there as well. We need only choose twoof the three possible loops: the top, bottom, or outer loops, but for practice we writeall 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 andgoing clockwise, we have

Make sure you understand why the signs are as they are (these are not arbitrary).Around the top loop, starting at the upper left corner and still going clockwise(note: the direction is arbitrary, but it is perhaps a good idea always to “travel”around loops the same way to help reduce mistakes)

�12 V � I1 (2 kÆ) � (I1 � I2)(1 kÆ) � 0.

�12 V � I1(2 kÆ) � I2(3 kÆ) � 6 V � 0.

I1

I2

12 V 2 kΩ

3 kΩ

1 kΩ

6 V

I1 + I2

FIGURE 16.11 Multiloop circuit forExample 16.4. Do you see why theseresistors are not in series or parallelwith each other?

We can also consider simple electrical circuits that have two capacitors C1 andC2 connected either in series or in parallel to a battery as shown in Figure 16.12. Asjust 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: thesame charge will flow from the battery, storing the same amount of potential energyas in the original situation with two capacitors. In the next section we show theeffects of having both resistors and capacitors in the same circuit, but first we com-plete this section by calculating the equivalent capacitance corresponding to thoseequations 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 eachcapacitor. Now, consider the portion of the circuit outlined in the dotted lines. Thissection of the circuit is completely isolated electrically and if it was originally neu-tral must remain so. Therefore the net negative charge on the right plate of C1 and thenet 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 ofthe voltages V1 and V2 across each capacitor and we have that

(16.15)

where Q is the common charge on each capacitor. The battery supplies positivecharge Q to the left plate of C1 which then induces an equal negative charge on itsadjoining right plate, resulting in an equal and opposite positive charge at the leftplate of C2 and an induced equal negative charge on its right plate. We show in thenext section that this “charging” of the capacitors when first connected to a batterytakes 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 capacitorwith capacitance C, that in order to have the same charge stored on this capacitor werequire that

(16.16)

Capacitors in series combine reciprocally, just as resistors in paralleldo according to Equation (16.14).

Using a similar analysis for capacitors in parallel, we see fromthe right-hand portion of Figure 16.12 that we now have that the totalcharge Q supplied by the battery is the sum of the charges on bothcapacitors: Q � Q1 � Q2. From this, we can write

(16.17)Q � Q1 � Q2 � C1 V1 � C2 V2,

V �Q

C�

Q

C1�

Q

C2 or 1

C�

1

C1�

1

C2. (capacitors in series)

V � V1 � V2 �Q

C1�

Q

C2,


Finally, although not needed, around the bottom loop, again clockwise from thelower left corner,

Now, picking any two of these three equations, we need to do the algebra tosolve for the two unknowns. We find that I1 � 3.82 mA and I2 � 0.55 mA.Check this for yourself.

� ( I1 � I2)(1 kÆ) � I2(3 kÆ) � 6 V � 0.

V

C1 C2

V1 V2

V

C2

C1

FIGURE 16.12 Two capacitors in a(left) simple series or (right) parallelcombination.

and again replacing the two capacitors with a single capacitor C and notingthat the voltages across each capacitor are the same because they are in paral-lel (V1 � V2 � V ), we find

(16.18)

Remember that, just as for resistors, these results for combining twocapacitors in series or parallel can easily be generalized to larger arrays ofcapacitors using the same tools as in the above discussion. Circuits withonly resistors or only capacitors present are ideals. In the next section weturn to a presentation of more realistic circuits with both resistors andcapacitors present. Such circuits are more realistic because there is alwaysa small amount of resistance (in the conducting wires themselves) or straycapacitance (between different conducting surfaces) present in any circuitregardless of whether an actual resistor or capacitor device is present inthe 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 witha specific capacitance (capacitance per unit area) of about 1 �F/cm2.This turns out to be a very good approximation for a pure phospholipidbilayer which has an extremely high resistivity of about 1015 -cm,comparable to a very good insulator. The very high equivalent resis-tance prevents charge from crossing the lipid region and maintains thestored charge as if the bilayer were an ideal capacitor. However, as dis-cussed in the last chapter, biological membranes are full of proteins thatact as channels allowing ionic currents to flow across a membrane.

The simplest model, or equivalent circuit, for a biological mem-brane in the resting state is shown in Figure 16.13 and is known as anRC series circuit. For now, we ignore how the equivalent capacitor wascharged (to a voltage V0 � Q0/C) and we imagine that at time zero theswitch S is closed (corresponding to the membrane channels opening),discharging the capacitor. The capacitor does not discharge instanta-neously, but follows a time course that depends on the values of R and C.The resistance R represents the effective resistance to current flowacross the membrane and is discussed further below.

To analyze this circuit, we use Kirchhoff’s loop method, discussedin the last section. Let’s write a loop equation for the circuit in Figure16.13 after the switch is closed and a path is provided for current flow.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 goingclockwise around the loop, we find

(16.19)

Because both Q and I vary with time, it turns out that we need calculus to solve thisequation (see box) to find that the charge on the capacitor and the current through theresistor are given by

(16.20a)

(16.20b)I � I0 e�

t

RC,

Q � Q0 e�

t

RC,

� IR �Q

C� 0.

C � C1 � C2. (capacitors in parallel)

Q � CV � C1 V � C2 V or


R

C

Qo–Qo

S

Starting from Equation (16.19), and substi-tuting from the definition of

(the minus sign is needed to make the cur-rent positive because it is equal to the timerate of decrease of the capacitor charge), theequation becomes

.

Rewriting, we have

.

Integrating both sides of this equation fromt � 0 to time t and from Q(t � 0) � Q0 to avalue of Q(t), written simply as Q, we find

,

so that

.

Taking the antilog of both sides, rememberingthat these logarithms are to the base e, we find

,

or Equation (16.20a). To then find the cur-rent as a function of time, we again use itsdefinition, so that

,

or Equation (16.20b). This same procedurecan be used to analyze any electrical circuitconsisting of batteries, capacitors, andresistors via the loop equation.

I � �dQ

dt� � Q0

d(e�

t

RC )

dt�

Q0

RCe

�t

RC

Q

Q0� e

�t

RC

log Q � log Q0 � logQ

Q0��

t

RC

LQ0

QdQ

Q��

1

RCL

t

0

dt

dQ

Q� �

dt

RC

RdQ

dt�

Q

C� 0

I � �dQ

dt

FIGURE 16.13 An RC series circuitwith the capacitor initially chargedbefore closing the switch Sconnected to the resistor.

where Q0 is the initial charge on the capacitor and I0 isthe initial current when the switch is closed and given byI0 � Q0/RC.

The results obtained in Equations (16.20) are shownin Figure 16.14 with the charge and current plotted asfunctions of time. Because the voltage across the capaci-tor is proportional to the charge (V � Q/C) and the volt-age across the resistor is also proportional to the current(V � IR), these voltages follow the same time courses asQ and I, respectively. The key parameter in these results isthe 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 thecharge, current, or voltage across either R or C dropping to (1/e) � 0.37 of its initialvalue in a time � � RC (see Figure 16.14).

All electrical devices and complete circuits have some associated capacitance aswell as resistance. In high-speed electrical applications, such as computers, the RCtime constant sets fundamental limits on the speed at which a circuit can change itsvoltage. Computers use voltage as information, with a high or low voltage represent-ing a bit of information, either a 1 or a 0, and calculations are done by electronicarithmetic that changes bits rapidly. Consequently, increasing the processing speed ofa computer depends heavily on reducing the associated capacitance of the funda-mental electronic device building blocks of the microprocessor.

M E M B R A N E E L E C T R I C A L C U R R E N T S 413

0

0.2

0.4

0.6

0.8

1

0 10 20 30 40 50 60time (ms)

No

rmal

ized

Q o

r I

FIGURE 16.14 Normalized capaci-tor charge or electric current in anRC circuit (Equations (16.20a) and(16.20b), normalized to their initialvalues) for a � � 15 ms RC timeconstant. The voltages across thecapacitor and resistor also followthe same time course. Dashed linesindicate that at t � �, the normal-ized Q or I has decreased to (1/e) �0.37 of its starting value of 1.0.

Example 16.5 In the simple RC circuit of Figure 16.13, the 10 �F capacitoris initially charged to 60 �C. When the switch is closed, an initial current of0.3 mA is measured in the circuit. Find the charge on the capacitor and thecurrent in the circuit after 0.6 s.

Solution: To learn the time course of the current and charge, we need to first findthe value of the resistance in the circuit. When the switch is first closed, the ini-tial 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, immediatelyproduces 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) tofind the charge and current after 0.6 s, equal to three time constants. Substitutingthat (t/RC) � 3, we find that the exponential is given by e�3 � 0.05, so that after0.6 s there will remain only 0.05 times the initial charge and current. Ouranswers then are that after 0.6 s there remain (60 �C)(0.05) � 3 �C of chargeand the current is (0.3 mA)(0.05) � 15 �A.

Let’s now apply some of these ideas to a biological membrane where we are par-ticularly interested in the transverse currents across the membrane. For membranesin the resting state, RC time constants range from 10 �s to 1 s. In dealing with mem-branes it is useful to discuss the electrical properties of a 1 cm2 area; these are knownas the specific capacitance C/A, and specific resistance RA. Defined in this way theproduct of the specific capacitance and specific resistance (C/A)(RA) � RC is stillequal 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 previouschapter of 1 �F/cm2, the different time constants correspond to different values forthe specific resistance RA � L of 10 to 106 -cm2. The broad range of values forthe resistivity indicates a large variability in both the numbers of channels per unitarea and in the average number of open channels in the resting state in different cells.

We now want to get some estimate of the numbers of charges flowing througheach open channel that make up the membrane current. Using a value of 0.1 V for theresting potential, we determined in the last chapter that a typical value for the surfacecharge density Q0/A is about 0.1 �C/cm2. Because 1 mol of a monovalent ion corre-sponds to a charge of , where Fis known as the Faraday constant, we can find the number of moles corresponding toa charge Q0 per unit area. If due to monovalent ions, the surface charge density cor-responds to

If we approximate the average current density (current per unit area) by dividing thecharge density value by the time constant, we find a current density I/A of100 �A/cm2 using a 1 ms time constant. This corresponds to the flow of 1 nmol ofions/cm2/s. Using a value of about 10 channels/�m2 (or 109 channels/cm2) for thesurface density of channels, the ratio of I/A (10�4 A/cm2) to channels/cm2 gives avalue for the current in a single channel of about 0.1 pA, corresponding to the flow ofabout 10�18 mol of ions/s. This means that each channel carries about 600,000 ions/sor about 600 ions in the 1 ms time constant. Measured values for a variety of singlechannels give currents of this magnitude or 10–100 times larger (see Section 6). Notethat the number of ions flowing across the membrane is insignificant in terms of thetotal concentrations of ions both in the cytoplasm and extracellular medium, so thatthe 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? Weshow that the selective permeability of the membrane to various ions, controlled bythe channels, is the source of this potential.

Suppose first that there are only K� channels in a membrane so that, to a goodapproximation, only those ions can cross the membrane barrier. If we start with anexcess of KCl on one side of the membrane, the K� will reach an equilibrium acrossthe membrane in which there is no net flow of ions even though the K� concentra-tion is not equal on both sides of the membrane. Why is this? Clearly in the absenceof any electrical effects, diffusion alone would tend to drive the K� concentration tothe same final value on both sides of the membrane. However, despite this diffusionaldriving force, electrical attractive forces due to the presence of the excess (negative Cl�)ions, which cannot cross the membrane, balance this tendency toward a uniformconcentration at equilibrium (Figure 16.15).

From an equilibrium equation similar to that of the discussion of Figure 13.6 inChapter 13, we can write that

(16.21)

where R is the molar gas constant, and the c’s and PE’s are molar concentrationsand potential energies, respectively, of the K� on the outside (o) and inside (i)of the membrane. Writing that where NA is Avogadro’s number, z is the valence or number of charges per ion(so that is the charge of a mole of ions), and VK is the equilibrium membranepotential due to potassium ions. Solving for VK by taking the natural logarithmof Equation (16.21), we have

(16.22)VK �RT

zF log a c0

cib .

zF

PE0 � PEi � NAq¢V � z F(V0 � V i) � zFVK,

c0

ci� e

�PE0�PEi

RT ,

(10�7 C/cm2)/F� 10�12 mol/cm2 � 1 p mol/cm2.

F � NA e � 6 � 1023 � 1.6 � 10�19L 105 C/mol


FIGURE 16.15 Portion of amembrane (with channels notshown) permeable only to K� (blue)showing that even at equilibrium,the concentration of K� is higheron the side with Cl� (pink) due toelectrical forces.

The Nernst potential represents the equilibrium situation for a particular ionspecies. If the transmembrane potential is equal to the Nernst potential for some ionspecies “A,” VA, then there will be no net flow of A across the membrane even if themembrane has a high conductivity for A. No net flow does not mean that the chan-nels do not allow any ion flow, but rather that the inward and outward flows of ion Aare equal. If the transmembrane potential is higher or lower than the Nernst potentialthen there will be a net flow of A one way or the other across the membrane with theionic current proportional to the difference between the actual potential and theNernst potential for that ion

(16.23)

where GA is the A ion conductance and V is the actual transmembrane potential. Ifonly the one ion species can cross the membrane, then the membrane potential willequilibrate at the Nernst potential for that ion. In the resting state, open K� channelsdominate and the resting potential is close to the equilibrium potential for K�, �0.1V. This behavior is identical to that expected if there were a battery in series with aresistor for each ion species. These separate batteries across the membrane functionwhen their corresponding channels are open, corresponding to when their series resis-tance decreases.

At this point in our discussion we can present a more realistic circuit diagramfor a membrane than a simple RC circuit. In the membranes of the axons of neu-rons, Na� and K� channels dominate, and Hodgkin and Huxley proposed theequivalent circuit shown in Figure 16.16. The arrows through the resistors in thefigure indicate conductances that can vary with time as the ionic channels are madeto open or close (known as gated channels). Only Na� and K� channels are explic-itly indicated with a net leakage conductance representing other net ion flows.Before we study some of the electrical properties of neurons and this equivalentcircuit representation in Section 5, we first give a more qualitative overview of thestructure and functioning of neurons and the ways in which their electricalproperties have been studied.

4. OVERVIEW OF NERVE STRUCTURE AND FUNCTION;MEASUREMENT TECHNIQUES

The human nervous system consists of some 1011 nerve cells, or neurons, eachone making an average of over 1000 interconnections. On an individual levelwe have a reasonable understanding of the functioning of a single nerve cell,

IA � GA (V � VA ),

OV E RV I E W O F N E RV E S T R U C T U R E A N D F U N C T I O N ; M E A S U R E M E N T T E C H N I Q U E S 415

Table 16.2 Typical Ion Concentrations and Nernst Potentials (Mammalian Skeletal Muscle)

Typical Internal Typical External Nernst PotentialIon Concentration (mM)* Concentration (mM) (mV)

Na� 12 145 �67

K� 155 4 �98

Ca2� 10�4 1.5 �129

Cl� 4 120 �90

* 1 mM � 10�3 M � 10�3 mol/L.

IC IK IL

outside

inside

C

VNa VK VL

GNa GK GL

INa

FIGURE 16.16 The Hodgkin–Huxley equivalent circuit for anaxon membrane. The batteriesrepresent the specific ion Nernstpotentials (L �� leakage, represent-ing the small contribution fromother ions), producing specificion currents as shown. The totalmembrane current is given by thesum of the four currents listedwith the capacitor current equal to(from Q �� CV)

where V is the voltage across themembrane.

IC � C¢V¢t

,

Equation (16.22) is known as the Nernst equation and determines the equilibriummembrane potential contribution from the imbalance of a particular ion, known as theNernst potential. Table 16.2 gives typical concentrations and Nernst potentials forNa�, K�, Ca2�, and Cl�.

but we have precious little knowledge of the larger-scale, or more global func-tioning, of our nervous system. Three main ways to categorize nerve cells includewhether they are part of the central (brain � spinal cord) or peripheral (all else)nervous systems, part of the autonomic (connections with involuntary muscles andinternal organs) or somatic (peripheral connections to voluntary muscles and sur-face sensors) nervous systems, or whether they are afferent (so-called sensoryneurons, carrying information from the peripheral to the central nervous system)or efferent (so-called motor neurons, carrying information in the opposite direc-tion). There are many different types of neurons, however, they all have commonfeatures and are believed to function in a very similar manner.

Neurons are single cells with a cell body containing a nucleus and usually asingle 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 awayfrom the cell body (Figure 16.17). Cell bodies tend to be clustered together inregions connected by bundles of axons. At the far end of the axon are the termi-nal endings.

Nerve cells conduct an electrical signal called the action potential, or nerveimpulse, discussed in detail in the next section. These signals are very similar in allnerves, traveling from the dendritic end to the terminal bundle end at speeds of up


Cell body

axon

Myelin sheath

dendrites

Terminalendings

FIGURE 16.17 Structure of theneuron (top) schematic; bottommultiphoton scanning microscopyview of nerve bundles (green) and aretinal “starburst” cell (red) found invisual processing network.

to 100 m/s. Usually each neuron is electrically isolated from the next and signals arepassed on to the next cell chemically. This occurs through the release of a neuro-transmitter from synaptic vesicles at the terminal endings. These chemicals diffuseacross the synapse, a small cleft between the terminal endings of one neuron and thedendrites of the next, and are detected by membrane receptors on the dendrites toprovoke an electrical response. Receptors are membrane bound proteins that, onbinding neurotransmitters either directly (through so-called ligand-gated channels)or indirectly through open ion channels, cause a membrane depolarization and acontinuation of the action potential. In certain neurons direct electrical connectionsbetween neighboring cells occur via “gap junctions,” pores connecting two neigh-boring cells that allow the direct passage of very small molecules. These are com-monly found in embryo tissue and are believed to provide a means for cell–cellcommunication in undeveloped tissue. In nerve cells, however, gap junctions do notallow as great a variety of control mechanisms as chemical synapses do, and aretherefore relatively rare.

It is useful to describe the overall circuitry involved in a simple reflex response. Ata minimum such a response requires four cells. The knee jerk reflex is well known asa 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 attachedmuscle is stretched. A stretch receptor senses this and produces an electrical responsethat is carried by an action potential along a sensory neuron to the spinal cord. There areflex response is generated as an action potential in a motor neuron returning to thesame muscle fiber. Arrival of this action potential generates a sequence of chemicalsteps that result in the contraction of the muscle, and the knee jerk response. A similarsequence of events occurs when you respond to a pinprick on your finger (Figure16.18). Of course this is a simplistic view, and there are other neural connections thatallow control over the sensory and motor signals from the central nervous system aswell, 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 usinginserted microelectrodes. Most of the early research work was done using thegiant axon from a squid, a particularly large cell with an axon of about 1 mm indiameter. The electrode is a glass capillary tube containing a conducting salt solu-tion and a metal wire electrode. Electrodes are used both to measure membranevoltages (with the wire inside the tube connected to a sensitive voltmeter) and toinject 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 membranepotential, 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 secondelectrode as a stimulus and subsequent changes in potential are measured.Alternatively, a constant voltage step change could be applied, fixing the mem-brane potential, and the changes in current flow across the membrane measured.This method is known as the voltage-clamp technique.

On first thought, one might guess that the membranecould be voltage-clamped by connecting an ideal batteryacross its thickness. The battery would supply whatevercurrent was needed to offset the membrane currents inorder to maintain a fixed membrane potential. This is, how-ever, not quite true because the battery terminals cannot be“attached” to the membrane and there are unpredictablejunction potentials at the metal–solution boundary due tocontact resistance that would vary with the current flow.Only the metal electrodes would be voltage-clamped, notthe membrane itself. Instead, voltage-clamping involvesusing an electric feedback loop to continually inject smallcurrents in order to maintain a fixed potential.

OV E RV I E W O F N E RV E S T R U C T U R E A N D F U N C T I O N ; M E A S U R E M E N T T E C H N I Q U E S 417

Withdraw finger response

motor neuron

sensory neuron

interneurons

pin prick stimulus

FIGURE 16.18 A simple reflex circuit.

Figure 16.19 shows three examples of voltage-clamp circuitry using feedbackloops. In each method, the membrane potentials are “space-clamped” in such away as to have no spatial variation of potential. In two of these methods two elec-trodes are used, with one measuring the potential relative to a reference voltage setat the desired level. This voltage difference signal is then used to inject a currentthrough the second electrode to reduce the difference signal and maintain the volt-age clamp. Such a procedure is an example of negative feedback, in which an“error signal” is sent back to the source and used to make small corrections so asto restore a desired value of a variable. The space-clamping is achieved by eitherusing long intracellular electrodes or by using a small membrane area isolated byeither applying insulators in gaps dividing the membrane or by a patch-clamparrangement. Patch-clamping, developed in 1976, uses a micron-diameter pipettetip pressed against an intact cell with some suction applied to form a very tightseal on a microscopic area of membrane so that the resistance between the insideand outside solutions is many G (1 G � 109 ). Patch-clamping has led to a100-fold increase in the sensitivity of membrane current measurements (seeSection 6 below).

5. ELECTRICAL PROPERTIES OF NEURONS

When several electrodes are used to probe the spatial pattern of normal membranepotentials it is found that small cells have membrane electric potentials that are con-stant over their entire surface whereas larger cells, such as neurons, can have poten-tials that vary spatially as well as temporally. Although a small cell’s membrane canbe reasonably modeled by a simple single-loop circuit diagram, Figure 16.20, inwhich the membrane voltage and current values depend on time, but not on spatiallocation (a so-called lumped-parameter model), neurons cannot.

Modeling the electrical properties of a neuron requires a so-called distributed-parameter network. The simplest scheme for a neuron that leads to some usefulresults 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 thelength x. Here the inner and outer conductors represent the intracellu-lar and extracellular fluid. Each section of length �x along the cablehas per-unit-length values of membrane capacitance cM, conductivitygM, transverse (inner to outer) current Im, and inner and outer longi-tudinal resistance ri and r0, as well as inner and outer values for lon-gitudinal current along the axon Ii and I0, and voltage differenceacross the membrane VM. The model was first developed to representan electrical cable (hence the name) that leaks some current trans-versely across the insulation between the two co-axial conductors.Although the mathematics of this model is complex, it is based on a


ammeter

injected current

feedbackamplifierinput

measuredvoltage

feedbackamplifier

A

measuredvoltage

injected current

ammeter

input

input

measuredvoltage

ammeter

membrane patch

pipet tip

feedback

amplifier

A

FIGURE 16.19 Three types of voltage-clamps. From left to right: gap method with insu-lating dividers, double electrode method for cells, patch-clamp method for pieces(patches) of membrane.

C

G

V

Vmembrane

Imembrane

FIGURE 16.20 Equivalent circuitfor the membrane of a small cellwith no spatial variation in itselectrical parameters.

straightforward application of Kirchhoff’s rules. Here we are content with showing afew of the model’s predictions.

Two parameters of the model are needed: the RC (� C/G) time constant, given by

(16.24)

and the space constant given by

(16.25)

Note how the units work out in Equation (16.24), with both cM and gM per-unit-lengthconstants so that their ratio has time units, whereas in Equation (16.25) the per-unit-length constants combine to give units of distance. The time constant is a propertysolely of the membrane with typical values of several ms, whereas the space constantdepends also on the cell dimensions and geometry and has typical values of several mm.If a steady electric current is applied at one point (x � 0) along a neuron, the mem-brane voltage difference VM from the resting potential decreases exponentially alongthe axon in either direction according to

(16.26)

as shown in Figure 16.22. After a brief initial time when the current is applied, thisresult is time-independent because current is continually injected by the electrode toachieve a steady state.

If, on the other hand, a short pulse of current is injectedinto an axon at x � 0 at time zero, the model can be used to cal-culate the voltage response as a function of both position andtime. This situation corresponds to a typical stimulation of anerve or muscle membrane. Results for this model are plottedin two ways in Figure 16.23. On the left the spatial variation ofthe voltage response is shown for several different times (dif-ferent curves). At increasing times the response spreads outfrom x � 0, decreasing in amplitude at x � 0, but increasing inamplitude at other locations for a brief time. This is perhapsbetter shown in the figure on the right where the time-dependence is plotted at several different distances from x � 0(given in units of ). The voltage rises and then falls with anexponential tail. The peak can be seen to move to farther loca-tions at later times, but with a rapidly decreasing amplitude. If

VM � VM (0)e�

|x|l ,

l�1

1(ri � r0)gM

.

tM �cM

gM,

E L E C T R I C A L P R O P E RT I E S O F N E U R O N S 419

cΔxgΔx

cΔxgΔx

cΔxgΔx

VVV

Ii (x)

Im (x)VM (x,t)

VM (x + Δx,t)

Io (x)

riΔ(x)

roΔx roΔx

riΔ(x)Ii (x + Δx)

Inside

Outside

Im (x + Δx)

Io(x + Δx)

x + Δx x + 2Δx

Ii (x + 2Δx)

Io(x + 2Δx)

Im (x + 2Δx)

FIGURE 16.21 A cable model forthe electrical properties of themembrane of a nerve axon. Thereare two parallel conductors alongthe inner and outer surfaces withrepeated transmembrane circuitelements representing the localcurrent-voltage characteristics thatvary with position.

FIGURE 16.22 The spatial variationin the membrane voltage frommeasurements along axons stimu-lated by a small current from anelectrode at x � 0.

the potential changes are below a threshold value, this will be the only response ofthe membrane, a localized brief signal. Data on so-called miniature end-plate poten-tials, due to spontaneously released neurotransmitters, are accurately modeled by thecable model. On the other hand, if the potentials exceed a threshold value, then atotally 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 wavethat is the basis of all neural communication. The Hodgkin–Huxley (H-H) model is ageneralization of the cable model in which the cross-membrane elements of the cableare spelled out in detail. In place of a single conductance channel, H-H uses three suchpaths, for K�, Na�, and for other leakage currents, with the conductances for Na� andK� given as variable conductances (shown with arrows through their equivalent resis-tor values in Figure 16.16). This latter change makes the entire problem nonlinearbecause the conductances for Na� and K� are now themselves functions of bothmembrane voltage and time. From Equation (16.23), we see that the ionic currents willnow depend on the membrane voltage in some nonlinear way (with the exponent ofVM 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 mea-surements (eliminating the x-dependence, or the cable properties) that were also volt-age-clamped, allowing direct measurement of membrane currents. Individualmembrane currents due to Na� and K� were measured by a number of methods,including radioactive labeling of the salt ions, or using channel blockers, specific chem-icals that block, or shut off, only one type of ion channel. From numerous measure-ments of currents at specific membrane voltages, plots of the conductances of each typeof channel as functions of potential were obtained. With empirical equations for theseconductances, the H-H model can account for all of the features of an action potential.

Figure 16.24 shows the time-dependence of an action potential and the asso-ciated ionic conductances. The Na� conductance increasesafter a time delay relative to the potential, peaks with thepotential, and then falls off more rapidly. Again relative tothe potential, the K� conductance rises more slowly andpeaks after the fall of the potential. Although the H-H modelwas developed under space and voltage-clamped conditions,it can explain a large number of distinguishing features of anaction potential, including: (1) an all-or-nothing response,with a threshold value of membrane current, in which a fixedpulse shape propagates down an axon at a constant speed; (2)an absolute refractory period of time after the action poten-tial during which a second action potential cannot be elicited;(3) a relative refractory period of time during which a secondaction potential can only be elicited by an elevated currentlevel substantially beyond a lower threshold; (4) a specificstrength-duration relation giving the threshold current for


0

0.2

0.4

0.6

0.8

1

1.2

–4 –2 0 2 4x/λ

no

rmal

ized

Vm

no

rmal

ized

Vmt /τ = 1/16

t/τ = 1

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4t/τ

FIGURE 16.23 (left) Spatial dependence of spreading membrane voltage at various times(decreasing voltage curves at x � 0 correspond to t/� � 1/16, 1/8, 1/4, 1/2, and 1); (right)Time-dependence of membrane potential at various distances from the stimulus at x � 0(decreasing peak voltage curves are at x/ � 0.5, 0.75, 1.0, 1.5 and 2).

FIGURE 16.24 Membrane voltagechanges during an action potential(bold), together with sodium andpotassium ion conductancesacross the membrane.

different duration current pulses; and (5) accommodation, in which the membraneadjusts to a sufficiently slow increase in current without producing an actionpotential.

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 nerveconduction mechanism. All neurons are found in association with other nonneuralcells, known as supporting cells. In the central nervous system these are mostly glialcells and in the peripheral nervous system they are predominantly Schwann cells. Mostof these supporting cells provide a myelin sheath that surrounds the axons of neuronsin segments, or internodes, that have narrow gaps, known as nodes of Ranvier, at regu-lar intervals. The internodes are roughly 1 mm long, some 1000 times longer than thenodes (Figure 16.25), and substantially change the electrical properties of nerve con-duction. The myelin sheath provides a highly insulating layer effectively reducingmembrane currents, which are fairly well confined to the nodes where there is a muchhigher density of channels than in the internodes. Myelin also greatly increases thespace constant so that the membrane potential changes occurring at one node spreadover many nearby nodes. Thus, a membrane potential depolarization occurring at onenode caused by local membrane currents will rapidly appear at nearby nodes triggeringmembrane currents there as well. This type of signal propagation is known as saltatoryconduction (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 continuousfashion along the axon. Action potentials generated by saltatory conduction travel atmuch faster speeds (up to 100 m/s versus 20 m/s in squid giant axons) and myelinatedneurons also have much smaller diameters (20 �m versus 0.5 mm in squid giantaxons). A number of neuromuscular diseases, including multiple sclerosis (MS), affectthe myelin around axons.

6. MEMBRANE CHANNELS: PART II

In Part I of our discussion of membrane channels in the previous chapter, wefocused on the control and selectivity of voltage-gated ion channels. Now that wehave learned something about electrical circuits, we return to membrane channelsand discuss patch-clamp measurements of the electric currents through single chan-nels. Patch-clamping was mentioned at the end of Section 4 above as a means ofspace-clamping, or electrically isolating a patch of a cell membrane. Four typesof patch-clamps can be distinguished for use in recording the electrical activity ofsingle channels (Figure 16.26).

A micropipette tip pushed up against a cell membrane provides an initial lowresistance seal of about 50 M. By applying suction, a gigaseal (G seal) is thenobtained where the patch is isolated from its surroundings by a huge resistancevalue; 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 extra-cellular molecules supplied by the pipette. If the pipette tip is withdrawn pullingon the membrane, the membrane will rupture at the pipette tip edge producing aninside-out patch mode (B) if done in air or in the absence of divalent cations. Bythen immersing the inside-out patch in an external solution, one can control theion 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 RT I I 421

Nodes of RanvierAxon

Myelin FIGURE 16.25 Myelin sheathsurrounding the axon with regularlyspaced nodes of Ranvier.

Alternatively, after forming a gigaseal, an additional pulse of suction or voltagewill open the membrane, exposing the cytosol to the pipette contents. This allowswhole-cell recording (C) to occur in which the pipette can introduce ions or chemi-cals or even proteins into the cell. Small cells can be studied quite well using thismethod. For larger cells, pulling on the attached membrane, by surface tension, leadsto the formation of an outside-out patch (D), with the extracellular face of the mem-brane able to be immersed in an external solution.

Each of these methods results in a patch of membrane (possibly the entire smallcell) as a boundary between two controlled solutions, one within the pipette and oneexternal. At that point current flow across the membrane can be monitored by elec-trodes and amplifier circuitry. With only a few channels per square micron of mem-brane area, the sensitivity of the electronics is such that single channel recordings canbe made in which the flow of typically 5 picoamperes (1 pA � 10�12 A) of currentlasting typically 1 ms can be measured. Such a flow corresponds to about 30,000monovalent ions through a single channel in the membrane. Our ability to measuresuch 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 membranesurfaces discussed in the last section, one might guess that the single channel Na� cur-

rent 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 cur-rent 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 macro-scopic current and farther apart, on average, during smaller macroscopic currents.Effectively, with a large number of identical channels, the current pulses add togetherto give the continuous macroscopic current curve. An alternative way to consider this


FIGURE 16.26 Types of patch-clamps.

I (pA)

Time (ms)

0

1

FIGURE 16.27 Single channelrecording from a patch-clamp.Each step in current is the openingand closing of a single ion channel.

is that repeated depolarization of the same channel, under identical conditions, willelicit a seemingly random, different response each time; but when those responses aresummed, the average single channel current is found to mimic the macroscopic currentobserved 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 chan-nel current as a function of the membrane potential difference from the equilibriumpotential for that ion species (see Equation (16.23)). Figure 16.29 shows an exampleof data from a K� channel. The slope is the conductance for that channel under theexperimental conditions.

A simple model of ionic channels can be developed in which the channels existin only two possible states, closed (C) with zero conductance and open (O) with aconstant conductance. Figure 16.30 shows a hypothetical energy diagram for thismodel. Note that the energy levels might well depend on the membrane voltage.According to equilibrium thermodynamics, the ratio of the number of open to closedchannels is given by the Boltzmann factor as

(16.27)

The probability that a channel is open, P0, is given by the ratio

(16.28)

A plot of P0 versus (E0 � EC)/kBT is shown in Figure 16.31. For large negative val-ues of the abscissa, corresponding to a higher closed than open state energy, with thedifference large compared to thermal energies, all channels are open. In the oppositelimit of large positive values, all channels are closed. When the open and closedenergy values are equal, 50% of the channels are open. The energy levels of bothstates change in response to the membrane potential.

In a further refinement of this model, various mechanistic models of the energies ofthe two states can be assumed. For example, one simple scheme is to imagine a gatingmolecule 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

P0 �N0

N0 � NC�

N0

NC

N0

NC� 1

�e�(E0�EC)/kBT

e�(E0�EC)/kBT � 1

�1

1 � e(E0�EC)�kBT.

N0

NC� e�(E0�EC )/kB T.

M E M B R A N E C H A N N E L S : PA RT I I 423

Time (ms)

I (pA)

V (mV) I (pA)

Vm (mV)

100–100

20

–20

Energy

Eo

Ec

closed open

FIGURE 16.28 The upper voltage represents an applied depolarizing voltageclamp, and the three single channel recordings, top three I versus t curves,indicate a series of repeated measurements of current in synch with theapplied voltage pulse. When many repeated single channel recordings aresummed (bottom curve) the macroscopic aggregate current that results isfound to be the same as when measurements are made over a larger surfacearea to give a macroscopic current signal directly in a single measurement.

FIGURE 16.29 Current-voltage relations fora single K� channel. The conductance ofthe channel can be obtained from the slopeof the dashed line.

FIGURE 16.30 Energy diagram fortwo-state model of an ion channel.

forces—controls the conductance of the channel (Figure 16.32). The interaction energycan then be written in terms of the dipole and the membrane potential and an analysisand comparison with data can lead to an estimate of the valence of the gating charge zeon the dipole. Hodgkin and Huxley’s work showed that z is about 6 for the sodium chan-nel, so that six positive charges are needed to shift from the cytosolic to the extracellu-lar side of the membrane in order to give the observed voltage-dependence for thegating. Equivalently 6 negative charges can shift across the membrane in the oppositedirection, or 12 charges could shift halfway across, and so on. Although some featuresof the H-H model can be recovered from this simple model, multistate channel models,with additional parameters, have also been developed.


+-

+

-

closed

open

FIGURE 16.32 A model for a two-state dipole control of channelopening and closing.

0

0.2

0.4

0.6

0.8

1

–10 –5 0 5 10(Eo-Ec)/kT

Po

FIGURE 16.31 Probability of channel beingopen as a function of the energy difference.

CHAPTER SUMMARYWhen placed in an electric field directed along a wire,the free electrons in the conductor move randomlyabout at thermal velocities (~106 m/s) while driftingalong the wire at very low speeds (~mm/s). The driftvelocity produces a net flow of charge Q, making up anelectric current I, defined as

(16.4)

In a conductor, the current is proportional to theapplied potential difference V through Ohm’s law

(16.7)

where R is the electrical resistance of the conductor. Ascurrent flows through a conductor, electrical energy is lostthermal energy through joule heating at a rate given by

(16.10)

Electrical circuits can be analyzed using two fundamentalrules: Kirchoff’s loop equation, stating that the net voltagedifference around any closed loop in a circuit is zero, and

P � IV � I2 R �V 2

R.

V � IR,

I �¢Q

¢t.

the junction rule, stating that at any branch point in a cir-cuit the total current into the branch point must equal thetotal current flowing out. In simple circuits with either justtwo resistors or two capacitors, we can develop the fol-lowing rules for finding net R and net C values:

(16.14)

(16.16)

(16.18)

In circuits with series R and C elements, an analysisfinds that when discharging the capacitor the charge onthe capacitor and the current in the circuit bothdecrease exponentially according to

(16.20a)Q � Q0 e�

t

RC,

C � C1 � C2 (capacitors in parallel).

1

C�

1

C1�

1

C2 (capacitors in series)

1

Requiv�

1

R1�

1

R2, ( resistors in parallel)

Requiv � R 1 � R 2 (resistors in series)

Q U E S T I O N S /P R O B L E M S 425

QUESTIONS1. Some mistakes students make as they are learning

about circuits involve using wrong language todescribe situations, leading to or caused by concep-tual misunderstandings. For example, it is fairly com-mon to hear students say that current flows across aresistor, or that voltage flows around the circuit, or toask what the voltage of the resistor is. What is wrongwith each of these statements?

2. Explain how it is that when you turn on an electriclight switch, the light comes on immediately eventhough the electrons making up the electric currenttravel at very slow speeds of only about mm/s.Develop an analogy with water coming out of a fullhose when the valve is first opened.

3. If conductors cannot have electrostatic fields withinthem, what is the mechanism that produces the forceon electrons within a conducting wire in a circuitwhen there is an electric current flowing?

4. If free electrons in a conducting wire experience a netforce due to the electric field in the wire, why don’tthey accelerate continuously instead of travelingalong with a constant average velocity?

5. Explain the difference among resistivity, resistance,conductivity, and conductance. Which are intrinsicproperties of a material and which depend on its sizeand shape?

6. If a resistor of resistance R is connected to a batteryof voltage V, the equation for the power dissipated inthe resistor, P � I2R, implies that a larger resistor willdissipate more energy and get hotter than a smallerresistor. This is not true. Explain why not.

7. A homeowner keeps losing electric power during ahot summer evening due to blown 20 A fuses. Afterreplacing the blown fuse several times and having thesame problem, he decides to use a 30 A fuse so itwon’t blow with the same electrical devices on. Whyis this a bad idea?

8. Copper wires covered with rubber-based insulation arecommonly used in household electrical wiring. Thesewires come in different gauges, corresponding to differ-ent diameters of the copper wire, where increasinggauge corresponds to decreasing diameter. Calculate theresistance of 100 m length of 14 gauge wire (1.63 mmdiameter) and of 10 gauge wire (2.59 mm diameter).According to National Electric Code standards, themaximum current capacities of these two wires are 15 Aand 25 A. Which can carry more current?

9. Check that the SI units of the product of R and C areseconds; verify that specific resistance times specificcapacitance also has units of seconds.

10. Two physics students are each measuring the RCtime constant of a simple series RC circuit. One ofthem sets the initial voltage on the capacitor to 10 Vand measures the time for the voltage to drop to 5 V.The second student, using the same circuit, sets theinitial voltage to 20 V and measures the time for thevoltage to drop to 10 V. Will these times be thesame? Why?

11. In circuit analysis, Kirchhoff’s loop equation is oftenequated with conservation of energy, whereasKirchhoff’s branching equation for currents is oftenequated with conservation of electric charge. Discussthis statement.

constant. Using this basic idea, nerve impulses can bemodeled as combinations of such potential differencesthat have different time behaviors in the Hodgkin–Huxleyand other models of nerve conduction.

To test these models, special techniques have beendeveloped to measure electrical properties of membranesin order to relate them to their structure. In particular,patch-clamping techniques allow scientists to study theelectrical properties of single special channels or pores,made from individual proteins embedded in biologicalmembranes, by measuring pulses of current flow corre-sponding to the opening of a single one of these mem-brane channels. The time-varying voltage signal seenfrom earlier voltage clamp measurements agrees withboth patch clamp measurements averaging over largenumbers of such pores, and with time-averaging singlecurrent pulse measurements after repeated channel open-ings, as seen, for example, in Figure 16.28.

(16.20b)

An analysis of the potential difference across a mem-brane with an imbalance in the concentration of ions onboth sides of the membrane leads to an equation, theNernst equation, for the potential difference across themembrane due to the difference in ion concentration onthe inside (ci) versus outside (c0)

(16.22)

where K is the example of potassium ions, R is the molargas constant, z is the ion valence and F is the Faraday

VK �RT

zF log a c0

cib

I � I0 e�

t

RC,

12. Explain in words what an equivalent resistor meanswhen replacing some collection of resistors in a cir-cuit by an equivalent resistor.

13. Explain why when using a multimeter as a voltmeterits two wire leads can be simply put in parallel, oracross, the circuit element whose voltage is to bemeasured, but when used as an ammeter this cannotbe done, but rather a wire leading to that circuit ele-ment must be “broken” so that the ammeter can beinserted in series with it. Discuss this in words and interms of Kirchhoff’s equations.

14. A flashlight bulb acts as a small resistance when con-nected to a battery. If two identical bulbs are con-nected in parallel to the same battery will they bebrighter, dimmer, or the same brightness as when asingle bulb is connected to that same battery? Repeatthis when the two bulbs are placed in series across thesame battery.

15. In the previous question, does the battery supplymore, less, or the same current with two bulbs in par-allel as when a single bulb is connected to the bat-tery? Answer this when the two bulbs are placed inseries across the battery.

16. Given an unlimited supply of 100 resistors, howcould you arrange a network of them to have anequivalent resistance of 150 ? Of 75 ?

17. Discuss the meaning of the two parameters of the lin-ear cable model of a neuron, the space and time con-stants. What do they tell us?

18. In the cable model, discuss in words the function ofeach 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 onone level, but are actually made up of discrete smallpackets on a finer level?

Questions 21 and 22 refer to: Consider the circuit to theright. The battery is a perfect source of emf. Treat A, B,C, and D as bulbs of equal resistance.

MULTIPLE CHOICE QUESTIONS1. 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 endand leaving at the �3 V end. (b) Electrons leaving theresistor have a higher kinetic energy than electrons enter-ing the resistor. (c) Electrons at the �3 V end have ahigher 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 equiv-alent (or effective) resistance of 2 . Which one ofthe following is true? (a) The sum of the individualresistances is 2 . (b) The sum of the reciprocals ofthe individual resistances is 2 . (c) Each of the indi-vidual resistances is greater than 2 . (d) Each of theindividual resistances is smaller than 2 .

3. The statement, “The current in a resistor is directlyproportional to the potential difference across theresistor,” is known as (a) Coulomb’s law, (b) Gauss’slaw, (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, becausethe wire’s volume doesn’t change, its cross-sectionalarea is halved. The electrical resistance of the stretchedwire is (a) R/2, (b) R, (c) 2R, (d) 4R.

5. A steady current flows through a resistor. An electronin 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 � kineticenergy, 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 differenceacross the poles of a battery. The battery is rated at20 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 reads20.0 V. When the switch is closed the ammeter reads1.0 A and the voltmeter reads 19.0 V. Which one ofthe following is most likely to be the explanation forthis result? (a) The ammeter has too little resistance.(b) The voltmeter has too much resistance. (c) Thebattery has an internal resistance of 1.0 . (d) Theresistor in the circuit has a resistance of 20.0 .

7. A battery has an emf of 10 V and an internal resis-tance of 1 . When the battery is connected to a com-bination of resistors, a perfect ammeter reads acurrent of 1 A leaving the positive pole of the battery.


A

B

DC

21. Rank order the brightness of the bulbs in the circuitshown.

22. Fill in the following table with “S” for same bright-ness, “D” for dimmer, “B” for brighter, and “O” forgoes out. “U” means that bulb is “unscrewed” for thatsituation.

Situation A B C D

1 U

2 U

3 U

4 U

12. V1 must read (a) 0 V, (b) 5 V, (c) 10 V, (d) some valuethat depends on the actual emf of the battery and theactual resistances.

13. V3 must read (a) 0 V, (b) 5 V, (c) 10 V, (d) some valuethat depends on the actual emf of the battery and theactual resistances.

14. A1 must read (a) 1 A, (b) 2 A, (c) 3 A, (d) some valuethat depends on the actual emf of the battery and theactual resistances.

15. A2 must read (a) 1 A, (b) 2 A, (c) 3 A, (d) some valuethat depends on the actual emf of the battery and theactual resistances.

16. Two light bulbs, one rated at 50 W and a second ratedat 100 W, are both supposed to be connected to a 110 Vsource of emf. Which one of the following is true? The50 W bulb has (a) twice the resistance as the 100 Wbulb, (b) four times the resistance of the 100 W bulb,(c) half as much resistance as the 100 W bulb, (d) onequarter 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, theswitch is closed. Which of the following best describesthe brightness of the bulb as a function of time?


31

2

8. Two resistors are in parallel as shown in the figureabove. When the current at point 1 is 0.15 A and thecurrent at point 2 is 0.05 A, what is the current atpoint 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 inseries. Resistor 1 has a potential difference across it of10 V and resistor 2 has a potential difference across itof 20 V. Now, the two resistors are connected to thesame battery in parallel. The potential differenceacross resistor 1 (a) is now 10 V, (b) is now 20 V,(c) is now 30 V, (d) cannot be calculated because theresistances aren’t given.

Questions 10 and 11 refer to the figure above10. 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 andeach resistor is 1 , what is the current through theresistor in the middle branch? (a) 3 A, (b) 2.4 A,(c) 1.2 A, (d) 0.6 A.

Brightness

Time

Brightness

Time

Brightness

Time

Brightness

Time

(a) (b)

(c) (d)

(a)

(c)

(b)

(d)

18. Which of the following best describes the potentialdifference across a capacitor that is connected in seriesto a resistor and a source of emf that is sequentially�V for time T, then 0 for time T, and so on, when T issmall compared with RC? Vertical axes are potentialdifference, horizontal axes are time. The lighter plotsare the emf, the bolder plots are the capacitor voltage.

At the same time, a perfect voltmeter placed acrossthe poles of the battery will read a potential differenceof (a) 11 V, (b) 10 V, (c) 9 V, (d) 1 V.

Questions 12–15 refer to the figure above. The A’s areperfect ammeters, the V’s are perfect voltmeters, the bat-tery is a perfect source of emf, and the resistors are equal.V2 reads 5 V and A3 reads 1 A.

PROBLEMS1. What is the equivalent current in a solution of

monovalent ions flowing through a capillarytube such that 1 mM of ions leaves the tube eachsecond.

2. A capacitor with a charge of 5 �C has its terminalsshorted by a metal wire so that the charge flows offwithin 2 �s. What is the average current flowing dur-ing that time?

3. What is the average current when all the sodium chan-nels on a 100 �m2 patch of muscle membrane opentogether for 1 ms? Assume a density of 50 sodiumchannels per �m2 of surface and a flow rate of 1000ions per ms through each channel.

4. Calculate the conductance and the resistance of a10 m length of 14 gauge copper wire, which has adiameter of 1.63 mm. If this wire is connecteddirectly to the terminals of a 12 V dc power supply,shorting it, how much current will flow assumingthe power supply can deliver an unlimited amountof current?

5. A 1 cm3 cube of gold ( � 1.61 � 10�8 m) isdrawn 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), oneof copper and one of aluminum, are soldered togetherand the 200 m wire is then connected to a 6 V dcpower 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, candischarge lethal currents of 1 A at 400 V. How muchpower does the eel generate?

9. An immersible heater coil is to be designed to heat aninsulated container with 4 liters of distilled waterfrom 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 resistanceto heat the water in 30 min or less?

10. Determine the equivalent resistance between points Aand B in the following circuit.

11. Given the network of equal 1 k resistors shownbelow, compute its equivalent resistance and the cur-rent drawn from the 12 V power supply. (Hint:Combine resistors in stages using the simple rules forseries and parallel combinations of resistors.)


A

B

5 kΩ 5 kΩ

4 kΩ1 kΩ

8 kΩ 3 kΩ

6 V 1.5 kΩ

4 kΩ

1.5 kΩ 2.5 kΩ

10 V1 kΩ

3 kΩ

1 kΩ 5 kΩ

V2

V1

10 kΩ

A2

A1

12. Analyze the circuit shown below to find the currentsflowing through and the power generated in eachresistor.

13. A single 1 M resistor is connected across a powersupply. An ammeter is inserted to measure the currentout of the battery. If a voltmeter with 10 M resis-tance is used to measure the voltage across theresistor, what will be the percent change in the currentreading on the ammeter when the voltmeter is con-nected across the resistor?

14. Find the reading that each of the (ideal) meters wouldhave in the following circuit.

15. Analyze the following circuit to find the current flow-ing through the 10 k resistor.

(c) Suppose the second identical resistor is placed inparallel with the first resistor, still connected tothe capacitor. What is the new time constant inthis case?

20. Consider a defibrillator, acting as a 32 �F capacitorand a 47 k resistor in a series RC circuit. The cir-cuitry in this system applies 5000 V to the RC circuitto 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 actingas plates with an air gap in between. Under certaincircumstances air can be made to conduct, so thatelectric charge can flow from the clouds to the groundin what we call a lightning bolt. Assuming that theclouds are distributed around the entire Earth at afixed distance of 5000 m above the ground of area4�R2

Earth, where REarth � 6400 km, the resistance ofthe air between the clouds and the ground is calcu-lated to be R � 300 .(a) Assume that the charge is distributed spherically,

so that V � k(Q/r) and therefore �V is the differ-ence in potential between the lower plate (theEarth’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 theclouds and the ground?

(b) What is the capacitance of the Earth–cloudcapacitor?

(c) If the charge on the clouds is discharged throughthe air, what is the capacitive time constant for thisdischarge?

(d) How many lightning strikes does this amount ofcharge correspond to if each lightning strike con-tains about 25 C of charge?

(e) Approximately how long would it take theEarth–cloud capacitor to discharge to 0.1% of itsinitial charge?

(f) Assuming that the charge is immediately replen-ished as soon as the discharge process ends,approximately how many lightning bolts are thereper day?

22. Fill in all the steps in the calculation of the number ofions crossing a membrane channel when it opens (seeSection 16.3 following Example 16.5). Now, usingthose same numbers, calculate the total number ofmoles of charge crossing the membrane when the


12 V5 kΩ

10 kΩ

6 V2 kΩ

16. Find the current in the central branch of the followingcircuit.

12 V3 kΩ

5 kΩ

12 V10 kΩ

17. RC time constants can be easily estimated by measuringthe time (known as the half-time) for the capacitor volt-age to decrease to half of some arbitrary starting valuewhen discharging through a resistor. From Equation(16.12a), the voltage across the capacitor will vary as

Show how a single measurement of the half-time canbe used to determine the RC time constant. (Hint:Substitute V(t) � V0/2.)

18. A 100 �F capacitor wired in a simple series RC cir-cuit is initially charged to 10 �C and then dischargedthrough 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 isflowing 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 con-stant of the circuit?

V(t) � V0 e�

t

RC.

membrane of a spherical cell with a radius of 10 �mcompletely depolarizes. If this charge were all K�

leaving the cell, calculate the fraction of the K� pre-sent in the cell interior that crosses the membranewhen it depolarizes. (Hint: You need to calculate thenumber of moles of K� inside the spherical cell andthe number on the surface of the membrane, all ofwhich is assumed to cross the membrane when itdepolarizes; see Table 16.2.)


23. Check the values of the Nernst potential in Table 16.2using Equation (16.22).

24. Show that Equation (16.28) leads to the plot shown inFigure 16.31. In particular, show that the P0 valuesfor E0 � EC and for large and small values of the dif-ference (E0 � EC) are correct.

Electric Current and Cell Membranes - if.ufrj.brcoelho/Newman/Newman16.pdfpotential difference...

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