882 CHAPTER 26 Direct-Current Circuits

known as an attenuator chain, since this chain of resistors causesthe potential difference between the upper and lower wires todecrease, or attenuate, along the length of the chain. (a) Show that ifthe potential difference between the points a and b in Fig. 26.91 is

then the potential difference between points c and d iswhere and the

total resistance of the network, is given in Challenge Problem26.91. (See the hint given in that problem.) (b) If the potential dif-ference between terminals a and b at the left end of the infinite net-work is show that the potential difference between the upperand lower wires n segments from the left end is

If how many segments are needed todecrease the potential difference to less than 1.0% of (c) Aninfinite attenuator chain provides a model of the propagation of avoltage pulse along a nerve fiber, or axon. Each segment of the net-work in Fig. P26.91 represents a short segment of the axon oflength The resistors represent the resistance of the fluidinside and outside the membrane wall of the axon. The resistanceof the membrane to current flowing through the wall is representedby For an axon segment of length

and (the membrane wallR2 = 8.0 * 108 ÆR1 = 6.4 * 103 Æ¢x = 1.0 mm,R2.

R1¢x.

V0?Vn

R1 = R2,Vn = V0>11 + b2n.

V0,

RT,b = 2R11RT + R22>RTR2Vcd = Vab>11 + b2,Vab,

is a good insulator). Calculate the total resistance and for aninfinitely long axon. (This is a good approximation, since thelength of an axon is much greater than its width; the largest axonsin the human nervous system are longer than 1 m but only about

in radius.) (d) By what fraction does the potential differ-ence between the inside and outside of the axon decrease over adistance of (e) The attenuation of the potential differencecalculated in part (d) shows that the axon cannot simply be a pas-sive, current-carrying electrical cable; the potential differencemust periodically be reinforced along the axon’s length. This rein-forcement mechanism is slow, so a signal propagates along theaxon at only about In situations where faster response isrequired, axons are covered with a segmented sheath of fattymyelin. The segments are about long, separated by gapscalled the nodes of Ranvier. The myelin increases the resistance ofa segment of the membrane to For such a myelinated axon, by what fraction does the potentialdifference between the inside and outside of the axon decreaseover the distance from one node of Ranvier to the next? Thissmaller attenuation means the propagation speed is increased.

3.3 * 1012 Æ.R2 =1.0-mm-long

2 mm

30 m>s.

2.0 mm?

10-7 m

bRT

Chapter Opening Question ?The potential difference V is the same across resistors connected inparallel. However, there is a different current I through each resis-tor if the resistances R are different:

Test Your Understanding Questions26.1 Answer: (a), (c), (d), (b) Here’s why: The three resistors inFig. 26.1a are in series, so In Fig. 26.1bthe three resistors are in parallel, so

and In Fig. 26.1c the second and thirdresistors are in parallel, so their equivalent resistance is givenby hence This combina-tion is in series with the first resistor, so the three resistors togetherhave equivalent resistance In Fig. 26.1dthe second and third resistors are in series, so their equivalent resist-ance is This combination is in parallel withthe first resistor, so the equivalent resistance of the three-resistorcombination is given by Hence

26.2 Answer: loop cbdac Equation (2) minus Eq. (1) gives

We can obtain this equation by applying the loop rule around thepath from c to b to d to a to c in Fig. 26.12. This isn’t a new equa-

- I211 Æ2 - 1I2 + I3212 Æ2 + 1I1 - I3211 Æ2 + I111 Æ2 = 0.

Req = 2R>3.1>Req = 1>R + 1>2R = 3>2R.

R23 = R + R = 2R.

Req = R + R>2 = 3R>2.

R23 = R>2.1>R23 = 1>R + 1>R = 2>R;R23

Req = R>3.1>R = 3>R1>R +1>Req = 1>R +

Req = R + R + R = 3R.

I = V>R.

tion, so it would not have helped with the solution of Example 26.6.26.3 Answers: (a) (ii), (b) (iii) An ammeter must always beplaced in series with the circuit element of interest, and a voltmetermust always be placed in parallel. Ideally the ammeter would havezero resistance and the voltmeter would have infinite resistance sothat their presence would have no effect on either the resistor cur-rent or the voltage. Neither of these idealizations is possible, butthe ammeter resistance should be much less than and the volt-meter resistance should be much greater than 26.4 Answer: (ii) After one time constant, and the initialcharge has decreased to

Hence the stored energy has decreased from toa fraction of its initial

value. This result doesn’t depend on the initial value of theenergy.26.5 Answer: no This is a very dangerous thing to do. The circuitbreaker will allow currents up to 40 A, double the rated value ofthe wiring. The amount of power dissipated in a sectionof wire can therefore be up to four times the rated value, so thewires could get very warm and start a fire.

Bridging ProblemAnswers: (a) 9.39 J (b) W (c) s(d) W7.43 * 103

4.65 * 10-42.02 * 104

P = I 2R

1>e2 = 0.1351Q0>e22>2C =Q02>2Ce2,

Q02>2CQ0>e.

Q0e-t/RC = Q0e-RC/RC = Q0e-1 =Q0

t = RC2 Æ.

2 Æ

Answers