Basic electric circuit theory a one semester text by i d mayergoyz wes lawson PDF

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ISBN-13: 97ß-0-12-48ß865-2 ISBN-10: 0-12-480865-4

Yuc rru! Jso ¢ontpktc your mjixsi œlirs via the Elicr ftoizepa¡p: Map:hawk. 0. Find (a) the voltage acrons the inductor, (b) tire power absorbed by the inductor, and (c) the energy storod in the inductor.

tb) Ni) — COi(i) = H o). Catculain tic nominal msictan‹:ri of tlu followiag auiomotive hardware (which run off a l2 V bsttery): (a) a 12 W donne light, tb) a 50 W horn, and (c) a 3fD W hmdlight. To rea‹ds a new customer, Oic lcx:aT electric compazty rttust tzaas ort five miles a cuzzeztt of /(I) = 20.1 cos(377a} amperes. What is the requiz¢d diameter of s copper wits djat would Losp the average power transmission losses below 2 LW? The car›duc‹iviiy of

copPer is = s.s x io’ (w•)". Aisume you have a carbon composition msisior with a oonductivirty of = CO X 1W(£lm) ' Cf you use this material tu make 2 cm long resistora. what is du requiied

diameter to produce a insistance of. (al J mât. why 1 ft, (c) t UN, and (d› i Mm which of theee design do yœi think arc prectical nues? How mighl you actua}ly construcl life

Cnleulate the résistance value for each of tne following cases of a cimularcrnis.settien

1&

a

'î

1

2

b

2 x III

2

I

CA pter 1. Basic Civil Veriabltc end Elevate f7.

Find the average{xwcr abcarfed by a 20 fif msisfor wk¢n a curzont of i(s) = 3/4 sin(3s) A is applied.

t&

A 10 pH indocio is driven by a curmnt i(i) = 3Dens(5(DI) mA. Calculate the power supplied tn the iralactor.

f9. TI›c voIt«gc across a 2 mH induc‹or is shown in Figuzc PI-19. Plot ‹he currsnc flowing uuough tI›c inducu›rusuming an initial cunent of 10 mA. t1 9

-1

0

T

2

9

4

5

6

7

8

9

10

H. The current mowing though a 39 pt-i induner is atiown in Figun Pt -2O. Plot the tenet in Inc inductor as a function of time. 10

2t.

A. Find expmssions The curmnt through a 91 JH inductor is given by i(i) = 7z*’ for the fluz linked by M inductor and thu voltage across the inducmr. What is the tolnl

ctaagc taai pacsm thmugh tne iixiuctor from t = 0sto i - ? Anume 9t0l = 0.

Caiculaic the capecitmxe for each of the folÍowing cases of a paralÍcl plale capacitor.

z.3

b

.I .0 l3

32.4

t0

Tue voltage across a 10 Wifi capacitor is shown in Figure PI-23. Plot the power supplied to the capacitor as a function of tir n

As

8

t0

t2

A 47 pF capacitat is driven by a collage v{i) = I + ó¥"“ V. Catcïil8tc the power supplied

The voltage across a 5 nF capacitor is given by Jz) = 1 + i + 3i° V for i > 0. Firui expressions for the curmnl, accumulated charge, instantaneous jxnver, and electriceneigy

siored in the capacitor.

The *oltage ac‹esa a 5 mH inductor is given by vti) = 3i° + Sr‘ V for I > 0. Find exp'ress’i o s for tlu current, magnetic flux, insiantaneous powur, and electric onetgy etored in tbn inductor. Let i(0) = 0. Thc current 0›rou8h a 33 vH i•ducia is given by /(i) = 20i cos(30rj n›A for I > 0. Find expressions for The voltage, magnetic Rux, instantaneous powcr, and electric enezgy sbM€d izt the indUcTOr.

32 H•

The Ctinuit through a 1 kfl

turismo ig green by i(i i - 2c ” sir•t iOf) mA fot i > O. Fitid

expressions for the voltage acrnss the resistor and tfie instantancous power disiipatea by 30. The voltege across a 1.4 fl rnsisini ix givm by vtiJ = e ”'° — e ”*^ V for i > 0. Find expressions for the current through the resistor and the instmitarieous Jxnver diseipaiod b tin resistor Plot tin ct dissi ated in the resistor

Chapter 2

Laws

2.1

Intmduction

Whai ezactiy is an electric cin:uit? So far circuit variables and circuit elements have been discussed, but we still have not defined precisely what an electric circuit is. basically, an electric circuit is an interconnection of circuit element. In this definition two things are important: the circuit elements themselves and the way in which they are connected. In circuit theory there exist iwo types of relaiionships between currents and voltagcs that correspond to the two facon of cleciric ci cuir. In the precious chapter, wc dcrived the terminal relations which ate deinnnined by the physical nature of the circuit elements. In this chaptcr we will present the rules thai govem the rami tications of the circuit interconriections. Thèse iElaiiorisliips 8ie

sometimes refermé to as topological oses. Before ptesenting thèse rulec we will first have to characterizc the topology (layout) of the circuit and introduœ the

tcrminology relaied to venons aspects of the circuit. Tne notion of the gmph of the electric circuit will be inooducea as a way to describe thn circuit inretconnections. Wc will ihcn present the two fundamental laws of circuit analysis, known at Kmi -hho ’s fnwr, that will be used in conjunction with the temiinal relations to determine the voltages and currents everywhere in a circuit. Afterward, we will demonstrate a method for assembling the propet number and types of mation thai will guarantn unique solutions for electric circuit problems. We will defer the discussion of the techniques needed to solve these systems of equaiions to the

following cbapieo. The chapter will close with the discuuion of resisiive circuits which contain only soumes and

Chepfer 2. Rizcfibo@s Leirs

2.3 Kirchhoff’s Laws 2-3.I Kirchhoffs Current Law Kirckhoff"scurrent law is generally abbreviated as KCL. /¢CLstatesthatt/tr algebraic um of electric currents ai »ode of an elactfi circt‹ii is equal io zem zu even

positive signs while others are taknn with negative signs. It is by con«nüon iliai we assign positive signs to the euwents entering the nodn and negative signs In the

curmnts leuving the nodc. Rnpressed mathemaiically, KCL is written as

(2.1) In cimuit theory, KCL is cnnsidercd to bc an axinm or postulatc. lt is tu bc accepted on faith and considered as a mathemaficd generalization of numerous

observations and experimental data. However, it should be noied that KCL can be proved by using electnmagnetic field theory, namely, the principle of continuity of electric current which was described in the last chapter. As an example. consider the ntxlc shown in Figure 2.4. The currents i and i3 are entering the node while the remaining currents are leaving th node. An application of equation (2.1) results in the expression: i — it -1- it — it — if = 0. With the aid DE tÏt8t

j3IiDCi@ÎC of CODtilïUi OU CICC C CUfYCnt, WC Cin îtÏS0 SCC

KCL applies io cut sets of a circuit. In this case, KCL would state that the algebraic sum of currents for any cm or is equal to zcra. However, in most cases discussed in this text it will be sufficient to consider only KCL for nodes. EXAMPLE it Consider the directed graph shown in Figure 2.5. INCL for node A reguizca that it + it — it = 0, XCL for node B require thet —I› — i = 0. KCL for node C yields it + is — it = 0, and KCL for node D yields i + it — i = 0.

Consider the surface S wttich cuu the circuit into two distinct parts via the fourth, sixib, and eigliih branches of the circuit. bc the branches conatituie

a eut ent.

Figure L4: A node with currents entering and leaving.

37

fiigure W KCL example.

Tire corresponding KCL equation is 0 — i# — i# = 0. The minus signs in front of the latter two terms arise because the currents from those branches are leaving the volume inside Sj. KCL yields i — i‹ - 0 for the cut set generated by the surface Sz indicated in the figure. There am two interesting points to note here. First, we can arrive at the KCL equation for a cut set by summing tire KCL equaiions for the nodes insidn the cut set. For flu surface Si, we sum the KCL equations for nodes A, B, and C and get it — i = 0 at claimed. Second, note that the negative sum of the twn cut set equations is equal to the KCL equation for node D (the only code which is not contained within a cut set). This example clearly demonstrates that the cut set equations follow from tln node equations (and vice versaJ and leads us io ilie notion of independent equations, which will be discussed laicr in this chapter. ■

ISO Kirchhoff’s Voltage Law Kiichhnff’s voltage law is the second fundamental axiom of circuit theory and is ofitn abbreviated as KVL. KVL is applied to voltages in loops and states that the algebmic satu of branch voltages amund any look of as electric circuit is eqml to ten at every detour of ii› e. Mathematically, it can be written as follows:

(22 To actually write KVL equations, we shall need the following nile for voltage polarities. We start by introducing reference voltgge polarities for each branch. fRe-

member that these reference voltages are coordinated wiui the reference currents for the passive elements.) Then, we trace each leop in an arbitrary direction (it is helptiil to trace every leap in the same direction let us always agree to gn clockwise in this text). If, while tracing through the loop, we cnkr the “plus" terminal nf an element and exit its “minus" terminal, then we take the voltage acmss this element with a positive sign. If, on the other hand, we enter the negaiivc terminal while tracing the loop, we take the corresponding voltage with a negative sign. To demonscau the above nile, consider the example circuit ahown in Figure 2.6

with the loops already labeled(keep in mind that these are nor the only possible leaps

Figure Nb: Example cix:uit with four loeps shown as I, O, III, IV.

for this circuit). KVL for loop I in Figure 2.6 yields —v * •3 + *2 ‘

0. Likewise,

for leop III we will get — vs — rs + = 0 ftom KVL. Since for passive elements, reference polarities for voltage and reference direc tions for current am coordinated, we can formulate the nile for determining the signs of brauch oltagcs in a XVL cquæion in tccms of ‹rte reference cuucnt directions. Namely, if tire teaming directinn coincides with the reference current direction, the branch voltage is taken with a positive sign in the KVL equation. Otherwise. the breach vnlmge mquirea a minus aign in the KVL equation. In leop 1, the tracing direction coincidcs with reference di‹•cti * *2 *2 Et ÎS Oÿ{fOSitC IO il. 50 we

would get -v, + vj + vt = 0 as we must Bernie th mference current dizr ciio and voltage polarities are not necessarily connoted for sources, one should afn'ays

we iW priors rule for voltage polarities to evat•aie the signof th vnJinges ocmss it sources in the KVL equation. EXAMPLE M Consider the circuit shnwn in Figure 2.7. Fuid all of the unknown voltages and currents. Solution: KVL for lms fIH IV) yisld, respectively: — 12 + vt + 15 = 0,

4 — 5 — vt = 0,

—vt + 2 — 4 = 0.

Figure L7: Example cimuit with tinknown currents and voltagm.

KCL for noden B, C, and D, respectively, yield: ia + 3.5 — d.75 = 0, is — 3.5 — ip — 0, 2 + 2 * ir = 0.

The first equation requires ia — 1.25 A and the ihird implies that ig = —4 A. Inserting this value into the secand equation completes thu solution of this problem by yie)dirg is = — I /2 A. KCL at node A can be used tocheck for algebraic mistakes. The resulting equation is 4.75 — 1.25 A — (—.5 A) — 2 A — 2 A =• 0, which ehecks out conectly.

2. 4 Li ne ar l yI nde pe nde ntKi r c hhoffEqua t à ons By writing equations according tn Kirchhoff’e laws. we end up with a system of linear algebraic equations. The task ie to guarantee that the resulting Kirchhoff equations am linearly indeed&•tii (i.e., solvablc and without redundant information). We will define linearly independent equations as those in whiCh each subsequent equation contains a variable (an unknown) which the previous equations in the system do not contain. Therefore, each equation cannot be obtained from the previous ones. In this sense, each equation cantatas esnntially new information. The given definition of linearly independent equations is a very special (particular) case of the general definition of linearly independent equations used linear algebra. However, the

2.4. Lirreztrty Jrufrprndrnf Birr

£§uo/iéxs

The variables Dn the right-hand side of the above equations can be used as sions

these variables. In order to achieve this, we substiiuk expressions (2.12) and (2.13) in the appropriate KVL equation (2.2) and expression t2.14) in the appropriate KCL equation (2.1). Once this is dune, rhc cult will be a system of linear differential equations which is written in terms of currents through the resistors and inductors and in terms of voltages acmss ibe capacitors. In this way, we reduce the total number nf unknowns to 6, which is the total number of equations. Thest diffemntial equations should be supplemented by the initial conditions for ip(i) and v •(i), respectively, ii*(0•)'

ii•(0-),

( 15)

vt•(0t) - vt•(0-1.

i2.16)

These initial conditions follow from the principles of continuity of electric cunents thmugb inductors and voltages across capacitor.

EXAMPLE M We shall illustrate the above discuuion with an example of an electric circuit, shown in Figure 2.12. Note that the source voltage dependence r (i) is assumed to in specified. We want to write b linearly independent equations with b unknowns. The first step is In specify the notles and to assign reference directions. Nene we shall write Kirchholf equations. By using KCL. we see thai the current

xConfornmksA,B ndC,m

uelym:

f()+i#)’i#)=0,

(217)

f(0-%()-i(O=0,

(2.18)

4(0*'(')-âV)=O.

(zis)

NDtiCfl that we do not write an equation for node D bnause ihis equation would not bt litteaflY ifide9cndent- N0tice arse that thefe is a trivial node E that connects nnly the capacitor and the negative terminal of the voltage snurcc. Rather than add another

R4

L s

Figure L12: Example timñt, with nodes and reference directions ahown....