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--!F-
fttria or Ef.ouc C}lbm
Syhmelrical componentsPosiliv6 sequenc€ componentsNegalive sequence componenrsZero sequ€ne conponentsGeneEJ lofiutas fo. symmeric€t
Per unit impedanc€ using a difte€ntporcr & voltage bass
The seougnce nelwo*Ths seqoence nslwoSingte tino to ground fautt
Double lins to ground fautt
Thee-phase t.utr €tcutanons gNen
Apprcximate eqlivatents belw€en aline to linefaLrtt& a 3-ph.s6fauli ot an untoaded gene.ator
TEST 24 (40 probtemstSOLUTIONS TO TEST 2'
696696696697
6S7
697698698698699700700701
702
749750750751
751
752752753
Tt4765
Th6oretrcat power ota wabr turbineSFcilic spe€d of a wat6r tulbineIu6ne generator sel upHeat en6rgy @nveEbn fictoBTems used in pEdiction ofload
Planl capacity taclor
TEST 26 (44 problems)SOLUTIONS TO TEST 26
TEST 27 (33 probtems)SOLUTIONS TO TEST 27
776784
7037U715
(two dimensionat ptane)
(lh€6 dhensionar ptan€)lrluminalion trom more than one
V.l!e of mountrng heighrto grvema)rmum lllumrnarron
Zonal cavity or av€rage tltuninat on
TEST 25 (20 probtems)SOLUTIONS TO TEST 25
735
736
736
737
f3?738743
SIRUGIUft OTI|IIIN
+ Matter - anything that occupi€s space and has woight
.:. Element- a substance lhat cannol be decomposed any fudher by chemical
.l Compouncl - a combination of lwo or more elemenls
.:. Molocule- smallest partrcle lhat s compound can be reduced to before ilbreaks down into its elements.
a Atom - smallest part lhal an element can be reduced to and slill keeping lheproperties of lhe element.
The atomic struclure ofan atom:
NuJ€ls "li. at"h -.?ite,.itli. il.ilit 6r'|.Etlt !,rc,!'ns i,]J,.,.ieutbnr .:md..l
<-a _irr. el<tbis revove.rDrn! jii o,bGor she 5alodid tLe i".cls
x..!s,
where: N = iotal number of electrcns on a given shelln: nrh shellofthe atom
9.107 x 1
1.672 x 10
!J!t!J!!!!!!!!e!!!4J!_!!!!!!t!!!Et!3e!ins b!, R Roras rt..l Valence electrons - electrons found in the outermost she or orbit ofan atom.
n Atomic Number - represents the number ofelectrons or protons of an alom
$ Atomic Mass represents the sum ofprotons and neulrons ofan atom.
3 da1rcn5 M lbe itt,a.te,j b\ ihe,l c ptarc anJ3 elR:tons drllb€ Ep.led irhc -rc;hr.mal,Dg r 6rllant hol,oi of6eecr oh5 qo nqrohad. rhe.r c Phte
The net num ber of electrons moved in the diredion olthe positive charge ptatedepends upon the porentjat difference between the two charges.
Electri.ity: Basic Principles 3
Volt (V) - unit of polenlial differencs, which is equalto onejoule of work donep.r ono coulomb ofchargs. Named afierthe ltaljan physicist, Atessand.o C.Voltr (1754 - 1827) who invented the firct electric battery.
tttclntcG[ $A body is said to be charge, if il has eirher an excess or deficit of etectrorc from itsnormalvalues due to sharing.
{. Coulomb (C)- unit ofetectric charge, which is equivatentto G.2S x t0r3electrons or protons. Named afrerlhe French physicist, Charles A. Coulomb(1736 - 1806).
?OTIXIIII IIITfIRTTSI
Potential- the capability of doing work
Any charge has lhe capabitily of doing work of movjng another charge either byattraclion or repulsion.
Example,A$ume1 cofcbarge.in movedi€lectron5.
l\\ _ 6+ F\I \\ (€+ l\\ betflohs* tbc:elcedbr'the,_ cpaea".r
". L,^n *_o'O-- l:) rer{rw,l b.ar,4drd b},--. cpL.e- l',.1 I <- TzLl I rarler,$.iar.rdoio.zet*,o. qor!
\ ll -oo-*\ ll 'o".'.rh",/.Da"\J o* \l
No+b. r \\ o+I 11 aj!+frcll o;\ ll o+\i o-+
Ampere (A) - unil of charge flow equalto one coulomb of charge past a givenpoint in ons s€cond. Naned affer lhe French physicist and mathematician,Andr6 M. Ampers (1775-1836).
ms$r cI
Ihe fact that a wire carrying a currenl can become hot, it is evident lhat the workdone by lh€ applied force in producing the current must b€ accomptished againstBome opposition or r€slstance.
+ Ohm (O) - praclical unit of r€sistsnce_ Named afi€r the ceman physicist,Georg S. Ohm (1787 - 1854).
TITCTIIC GUIITXT
Wh€n a polential difference between lwo charges forces a third charge to mov€, thecharuo in motion is called an el€ctric current.
where: R = €sistance (ohm)A = cross-sectional area (square meter)p = resislivity (ohm-meter)L = Ienglh (meter)V = volume cubic met€r)
.l Specific resistance (!$istivity) - resistance offered by a unil cube of the
.! Circularmil(CM) - area ofa circle having a diarneter (d) of ono rnit.
Ri*', R'
-r----
1 l00l Solrtd Ptublen$ in Elect cal Ensineeins by R. Rot.ts Jr.
cM-il.. 1,000 mil= I inchI McM = 1,000 CM
IEIGI 0I It Pmlrufit n$t$ tc[Experiments have shown lhal the resistance of all wires generally used inpraclice in elecl cal syslems, increases as the temperature increases.
r, = radi'rs otthe cabler, = radius 1o the ouler surface oflhe insulation
Elect citv: Basic Ptihcible.\ 5
G0x0tfitxcIConductance is a measure ofthe malerial's abilitv lo conducl electric cuffenl ltrs equalto lhe reciprocalof reaistance.
Slomons (fomerly mho) - unil of conductance. Named aflerthe Gemanengineer, Em6t Weme. von Siemens (181G1892)
R1= initial resistance (ohm)R, = linal resistance (ohm)T = inferred absolute temperalure
= temperature when resistance of a given materialis zerolr : initial tempe€ture12 = final lemperaturcAi = change in temperatureo = temperature coefficienl of resistance
6 = conductivity (siemens per meleoL = lensth (meler)A - cross-sectional area (square meler)p = specilic resislance (ohm meter)G = conduclance (siemens)R = resislance (ohm)
Tempe.ature coetficient of resislance (o)- ohmic change per degree per ohmat some specified temperature.
lffiultfl0r llst$txct 0r flflrs
@" rt:*'"fp = resisiivily of the insulating male al (ohm-mete0I = length of cable (melet
Ubs{ t.Problcnl t!
-.- I TS,ua19qg"9 Aietoctdc has a chsrge of 2 coutombs. tf 12.S x 1O1s treeeredrons are added to it, whatwillbe the net charge on lhe said dielectric?
A_ 4Ca. -2cc. 8cD. Zeto
Problqn ar
., Aiatt€ry.c€n detiver 't 0 joutes ofenergy to move 5 coutombs ofcharge. Whal rslhe potentiat difference betwe€n the termin;s ofti6 ba(6ry?
A.2VB. 50Vc. 0.5 vD.'5V
ProblGrn ttA clorid of 2.5 x 1Or0 elecfrons move pasl a given point ev€ry 2 seconds. Howmuch is the intensity of the €lectmn t!e?
a. 2Ac. 2.5 AD. 1.5 A
Problcltt 4:, The curenl in €n 6lectdc tamp ls S amperes. Whet quanlily of etoctricity fowstowards the litament in 6 minulesi
A. 30CB. 3600 Cc. 72CD. 1800 c
Eroblclo 5rA constant current ot 4 A charges a capacitor. How tong wilt it lake toaccumulale a tolalcharge of g coutomb; on the ptates?
B.c.D.
2s32s1na'12 s
Ploblcdr Or EE 8o.rd X|rch t99aThe 3ubstation bus bar is mad€ up of 2-iches round gopper bars 20 fl. long.
What is lh€ resislance of each bar if resistivlty iB 1.724 x 10'ohln-cm
A.7.21x1OsohmB. 13.8 x 106 ohmC.5.185x105ohmD.2.96x106ohm
Problcr! 7r EE !o.!ll bbcr r9y,D€lemine lhe rasislanc€ ot s bus bar made of copper if th6 l€ngth is j O meters
long and the crGs-section is a 4 x 4 cln2. Use f.i241 micro ;hm"ctn as the|esislivity.
A.2.121x1O{ohm8.4.312x1O'ohmC. 3-431 r 1o6ohhD. 1.078 x 1Or ohm
P.aobL!ANk$rome ribbon r$lstor elemenls 6ach ha6 a 63i3tenco of 1 ohm, The €lement
's made from sh€et of nichrome alloy, 0.m5 crn thick. tf ths width of th€ ribbon is 0.3
cm. what longth iB requlr€d p€r elcm6nt? Assurne specific €3istance of nidrrcmealloy to b€ 1m 4-cm.
A.68.8crnB. . 62.1 cmC. 70.7 cnD. 67.4 cm
PlobLnl 9r EE lo.r.d Oetota. t99tOne lum o{ a copper bar is produced by cutting a copp€r wash€r along a mdius
and spreading the end6. The washer is dI from soildrawn copper having aresistivily at 2O'C of l.732x1orohm-dn. The washer is 0.125 inch thick and hasinside diamele. and outside diameter ol 1 inch and g ind|eB r$pectivety. C6lculetethe exacl .€6islance betwoen the two en& of th€ tum to direct curent; taking intoaccount lhe non-uniiom qrnenl dislribution. Assume lhe contacl along the ends oflhe turn to b€ perfect ovff the 6ntire cross€eclion_
A. 12.74 x 10 6ohm
B. 15.53 x 10{ ohm
C. 17.22 x 1Or ohm
t00l 'olv.d
Prcbl.nt tn Et.ct cat
O. 14.83x 10{ onln
PaotL- tcr ED loaaa 6!.. taio1.0 #mf,*f::"*lr-mor ! condicior o.1o_ m tors. with . uniform drrm.rort.o c"',no rro"rne
"'."l*"ny-"irfr";;il; ;i;iffi:i Jffi;"fH"*ffon€ 6nd otfrr! condqdor accordiag lo fl. foffula: p - O.OOf + f O<f., otric,nA.0.0852ohm8. 0.0015ofimC.0.0806ohmD.0.0902otrn
htr ur trtlc.r{ ','Drllta},""iffi 8,H,,l;tdil?T; Hff :i,1ff"H"i*"#ilJ.#:"Ji
A. 1896 ohr|!g. 1825 ohmtC. 17gZ c'vnD. t8O5 ohrnr
#A. 4AB. 3rrc. 5rrD. 6n
Pr!t- lc- r!|t EA fo$d Octob., taca,",r"i'gfiH,H"*ft.i"1;:H'c' r!;rewn untir ft! resbt nce is 1oo tirn€" rhe
10nt00 m12.5 m5m
P.loblc.r 14, tE !o.rd Alstt tro:]," .#i':ffi :y#fi :?HH#'","Hl*';;lilT;:k ffl:ff s #,0#'*'.A. 0.85ohm
Teet I 9
1.0.?8ohmC, 0.e3 ohmD, O.01ohm
tiaDLrr ttA oo.ducto{ wttos€ diameter is 0.175 inch hAs a rosislsnce ot 0.5 ohm. Th€ wire
h dnwfl lhroogh a sedes of dies udit its dianr€ter is r€duced to 0.OS ind,. A;;;lta apaclfic rrristance ot the materiat r€main3 con"t nt, *t
"f ij ff," ,""i"L-J &
Itneih.ncd conducior?
A. l l.,l5 oa, 10.22 Ac. 12.75AD. 10.82 ()
tr.ll.rr t& EP !o.rd Apfll rgtjA corloin wir€ has a ro3ldiance R, The r$iltance of another wire fttenticel wilh
lha ll..l except for havlng twico its diemet€r is
A, 'Re. ln2
c. 2R
o. 1n1
trou.ttt t?rA givan wir6 ha6 a r6sbtance of 17.5 ohms. It its t€ngth b 560 m, how much
langth mwt b€ dd-off tom the r|i€ in order to reduce its reiistancs to ti.s otrms.-
A, 160 m8. 170 mC. t45 mD. 155 m
PlcoblCltr tA EE lo.|r! oglotcr t''6Whet is th€ stze jn squaB miltim€l€r (mmr) is the cabte of 260 MCM 6ize?
A 118.656 mm'?B. 126.675 ll,m2C. 112.509 mm'zD. 132.348 mm2
Pr.obl.tn l9rA 500 MCM ACSR cabt6 has 37 strands. Dete.mine the diameter in mils of each
8.c.D.
TO
B,c.D.
l00l Solvcd prcblens in Etectticat
116.25 .120.24118 34110.35
Prioblcd 2ot
;#i#f #fi ,::#rx$;"1a,:"Jiff ';Hj"1r,T:"["j,ilHri;rT, jt:
A. 800 ttllcM8.820.t\rcMc. 850 MCMD, 9OO MCM
ll#!.nitrffilfiffi fr'#iic"gr,""g;,g;y"r".,".,,.A.72.26ohms
3 3i3i3i$iD.58.15ohms
P|!obLa!ta,
fl rfli"l:H*:#$ffi 'ft1ir6r:q,f
{,:r{!ffirll,1##A. 88.89aB.90.12oc.85.22()D. 92.81fi
Problcnt A[
s:*l,!";li$"fl ..*;[i'$*ur;:"f:si#Hrrli*iiJlr""#JixA. 60.4.CB. 59.2.Cc. 58.4 "CD. 53.7.C
Test I l1
PtobLo t{l EE Eo.rd L.!Gh rt9!Th€ r68blenc6 of a wire is 126.48 ohms at 100 'C and 100 ohms at 30'C.
Dalalmlno lh€ l€mp€rature coefficient ol copper at 0 "C.
A, 0.00427I 0.00615c 0.02580 0 0356
trablod r3rTwo wire8 A & B made trom two diffgr9nl materials have temperature coeflicient
Cl r.rlslenc€ 6quel to 0.0025 and 0.0005 ohm per "C, respectively. tt is deGircd tomakc s coil of wire having a aesistenc€ of 1200 ohms \Nith a tomperature coefficientOt 0.001, u8ing suitabl€ length6 of the tvo given wires connecled in series.oalarmlne lhe required length of rvir6 A?
A, 5.5 mg. 8.2 mC 8.0 mD. 5.0 m
Ftobld! 16r EE lo.rd OctoDar tt tTs,o heating €lem€nls which is 500 ohrns and 25O ohms are connected in sedes
wllh l€mp€ratu€ co€ffci€nls of 0.001 and 0.m3 ohms per 'C, resp€dively at 20 .C.
Calcul5le lh€ effediva lomp€rature coefficient of lhe combination.
0.002150.006260.00712000167
ProbIod !7tA cylindrical rubber insulaled cable has a diameter of 0.18 tnch and an
lnlulation thickness of0.25 inch. lfthe specific resistance of rubber is 1014 ohm-cm.dclermine lhe insulalion Gsistance per 1000-fi length of the cabte.
A.624MOB. 682 MOc. 678 MOD. 694 Mc)
Problelr tarThe diameter of a given bare conduclor is 0.50 inch. A thermoptastic insulation
with lhickness of 0.1 inch is Mapped around lo jnsulate lhe conductor. Determinethe ansulation resistance of thjs conductor per meler. Assume specific resistance oflhermoplastic insulation to b€ 2 )( 1 014 ohm - cm .
c.D,
I2 l00l Solved '
A- 107xroP{)B. l(}5x'toeoC- llOxl0eoD 109x lOeA
PFoblc|! t9. tE to.!d Oclo0cs tgttTtte insulalion resistance of a kilo.neter of the c€bl6 haling a diameler of 2 cm
and an inG{dation tickoes ot 2.rn b 600 otms. It the thic*n€;s of the ansutation iginc.eas€d to 3 crn, find the in€ulatio,r r€6i6bnce otthe c€ble.
A. 725 ohms8. 850 ohmsC. 757 olmsD.828ohms
Pr.DLd tiot lE fc.a Oettb.r riaon€ a€si6k1ce ot 1m rneie{s otBire is 12 ohms. Wrat is its condl€ ance?
0.0521 mno0.0€i] mho6 mho6!2n lo3
I Notc volt-opulci ps@ul@b
o68.2 voll|
B.
-m2f8rrr?4
Solutons to Tesl I 13
i.tulrsbL$lOr . +2C
o? - -12.5 xlor'€ x;2i*1% = -2c
o,n - o, +O, -2+G2)Q* 'o
E::'ff:;,H"';"'
l - 2dfipgratt
E o-n-'(ur"'+#;a -18@c
o8t4
l-2d€c
^-€-a1= 2ex r 1319 r 2 9{ 9n = 649 6 o','ln 1n l!
^ t (1.724 x 1O3Xm96)25238
R -5.t85x105drt
'lrir,"r,r 2 Tl't lincfi /
1.D7. O !3_A t9_ A 25. C2.A 8. A t4_C 20. a 26. D3.8 9. B 15. A 21.C 27.O4.O 10.C 16.O n.A28.A5.411_A 17. A ?'3. B 29. C6. C 12. C 1A. a 2.t- A 30_ B
??-30 - Iopnotch€.
:!5-e0 - Condition.l
O-I{ - F€iled
H l00l Solved Prcblens in Elechical Enpineetine by R Ro,as Jt
L = 1om x 1991! = looo cm
1m
^ L {1-7241x 104 )(1ooo)' A (4x4)
R =107.75x 103 ohm
E L=!l=(Iqg!-9oL = 68.8 cm
*=oiComid€f a difiarential elenrotA=tdrL=2nt
=frn'1* =f.tr"',-r"1r
2a11.724x1041
G =6?902.25mho
-1 1
G U102.25R = 15.527 x 103 ()
"nd'?0?r,L=0.1m=10cm .
_L
(tnl.s- tno.s)
n . f, Jo aL = f Jro
oor.ro 'r:r +
". i[.-".'. 1'; ]]l' = i [. **'.''{S)]
R - 0.0806()
o.125inch x12 s4cm )
^ t l inch l
Sofutio^ to Test I 15
an = d4I)ldr
^1R
ce = lt*tJl2't \
c=-t fdt2olr
I
lr
I.I
L1
c::::33:::::5L,= 2.5L1
R1
1:::::::..:.-)&=100R)
I *-'$Nota Sinccthe coll ir roound with thc same weight orwke, thusthe vollnres ofth. wir. us.d in both coDditionr rre con,Qnt. Ard witl lhc volumc coDttaDt,
rcrttance lqfleJ drFctly a. the qua'e ofthe length.
R=kL2; k = proportionslity conBtant
n" tr"i'R' lL,JR- fbnorh o€r tumxtL \t ttsaool'380 ( bngfi p€r tum r t{ J \ 6000./
Rz = 1895 35O
Notc: Asruming thclc ar. no losscs in the prcc€si, tb€ volum€ ofthe mat€dal
rcmains constaDt And wlth the volume <on5tant, the fe3'fince ofthe nateial
'anesdrcctly a: ihe squa'e ofthc lcngth.
R = k L2; k = proportionality constant
R, TL, I,& -|.r.,]
"^ =*. l!.J' - o.o1z sr,
1'' lL,/ I L, .l
Rz =54
R = k L, ; k = pfoportionality constant
R, IL, ],R. =l\
J
, lE; . Foo R;
"="'1& = I &L, = 10m
16 l00l Solved Problens in Elect cal Sofutio,"t to Te.r I 17
n=oY= Pv -=rr'Ya' (1,\'14)
Notc: A5romiDg th.rc arc ro I65e5 jn thc proce$, thc voloD. of rlrrcrrainr .onstart And wtth tbc volume onstant, the rcsirtan.e ofthe*tre:nrcncly ar thc (ou,tfi powcrorthe d amerer.
R=*; k =proporlionatity constanl
!e=f4 fR' t d,.J
*, =".f 4, ]' -oo.r1 rr'zf' ldzl \5,R: = o.€3 o
R = *; * =proporlbnality con3tant
&=f d,)'R, 14 ,J
*=o[#)'=,'(#iR2-1145a
c=ol = 4 =r LAltdz
Note.Arlrm. th< lalgth of{'c two wkcj ir botb cordltlors to be conrtaDt Andwrtl thc lcngth cor5tant, tLc rerstance oftlc wrie latres rnveisely ar the square oi
R=-:i f = proportona$y con {ar
I Notc t Dgth
'D both @nditron! are tle eme
p. - !.!.;0. = tzo-cup.rn
R. =*ie. = 10.37 r! - clil per ff
R. =Rcp.L. _ p.L"
\,, -rJ!_r lL, Length
cut-of
-__.\
\E--._1, d/
J7-ttrand ACSR c6U.
R, ld,l'n: =l4l
n. =n.l4l'=^ll')'^ ' \d,I l2d, )
", =?o
I u'"1N'n. Alrrm. thc dott 5<tiooal arc! oFth. wrrc to be.orsbnt And with th.(n^r k1tr.r)nl ca conjtant, rc'iri.oc. ofth. wjre \€ri., drectly astle let]gth.
R -tti k - propodionality c.mlantR, t-R' L'
r. -!&.!9!l?9=1oo''' R, 17.5
Lamfi crrl -off = L -1, = 560- 400
l-anClh cu( - of = 160 m
26O MCM = 25O,@O CM
c -da. JEu " Jzsoooo , soo-1",
---ii$,2ll" - rz 2,,,n'- l0o0rn13 linch
^ d ann"14
A.l26.67mrn'?
t(b tlCM = 500,000 Cil{
r.- -!Q!99=1*rr.u,*cu. dd-Jet=Jfu5135d-116-25mils
.1,=2d'
;
E -="#=ffi="#
R, fd, l'R,,
= td,j
n, = n, 1!r l'
R- =:lo'1
Note: Aisumirg tlerc arc no lorse5 ir t6. procets, rhc volumc oF the rrateralieirairs constart. Ar.l with tlc voJum..oDnrn{ tbe resrsence ofthe materrallzrr.5 inv.6cly atthc Foudh power ofthe,ljamerd
_kr.(=-, t = proportionatity constanl
!.=ld' iR' ld, J
*, = ^.I
a, ]'- o 0.,1 rr'z I' (dr/ \5,R:=093o
R=f ; k = nrolortionalig conatant
R, Id,)'R, ldr.l
R, = R.l.dtl'=o5lo ris f- \d,) \ 0.06 lRz''11.45n
n=ol = 4 -r, L'A aA2 d,4
Note,.Ar5urrc th€ largtl ofthc two wr.s ir leth cor,ljtron, to be con5tant Ardwiih tlc lcngil, corrtant. tlc rerrsrarccoftlc nrcvariu: inu.oely as th"sq,a," oi
R=+; k - propondna&y corl3tafil
/___-t,
=Rl-!Ll\2d,j
t-
d, =2d'
Solutioti; to Test I 17
_L
Notc Anumc thc doss s.ctional area oFth. wrrc to be constnrt And wjfh tberkrs 5..tronilarca conrtant, retisiaD.r o{the wire v?do dfectlyasthe lenqth
R-kLi k = propodionality conslantR. L.R, L,
1^ =lEa= $o(12 5)
= a66 -' R, 17.5
Longth crrl - o{f = lr - L" = 560 - 40O
Longth cul - off = 160 m
250 MCM = 250,000 Ci'cM= dd = JcM - "65oJoo
. soo -r" , lind ,54tt -12.7.n.,'' l(mmtu litrn
. 'd d12n241
A = 126.67mm2
500 McM = 500,0m ca4
1.-,0 =!4@=rrra_srcucM=dd = Jct = Ji 3s13,sld = 1t6 25 mils
\____rl__,JL, Length
cut-olf
'7-strand ACSR caue
E Note, L$gth' ir boil ondltronr are tle 5ame
p" =!&;o. = rzo-cLrp.rt
p. = !t!r;0. = ro.:zn- cup.,a
p,L. _ P"L.
It 1001 Solved Ptublens in Electical
a = &e. = 49.!1) = ars ezz' p" 10.37
A. :820 cM
do - i- zgl:e6so . 0.00427
R, - R.(1+ 0.0025ar) - Rb(1+ 0.0005^, ) -OR, - (R. + RbXl+ 0.001a,) =O
Equato Eq.1toEq.2:R.(1+0.0025^r) + Rb(1+ 0.0005^,) = (R. + RbXl +0 001ai)
0.0015R.Ar =0.0005RoArRr = 3Ru
R. +Rb = 1200
Rr +3R. = 1200
R. = 300 o300
L.=6m
R, = 5OO(1 +O.0o1l') + 250(1 +0.003ar) = OR, = (500 + 250X1+ a* ai ) =A
Equate Eq.1toEq.2:
500 (1+ 0.001^r)+ 250(1+ 0.003a,) = (750X1+ od'^,)500 +0.5ar + 250 +0.75A( = 750 + 750o.nar.
dji =0.00107
E ""=+r = --a- =zu.tgz,c
0,00427
!e_ T+12R1 T+t1
- - /r*.-) 5o(234.1s2+1oo)..=\tfaj= ,3.rrr.3oRz - 63.24 o
r =1=-1-= 256"sdo O.o(X
Rr_T+t,Rj T +t1
R- = R. ffjlzl = so(250 * so)
' 'lT* t j 250+20
R, - 88 89o
1=1=-1-=zso"cao 0.004
R, _ T+t,R1 T+t!
r^ = ft*r.)rla)-r' \R,l= tzso * r stl.119) - zso' 't120,
r, = 59.2 "C
R, _ T+1,Rr T+ lr100 T+30
126,48 = T-1ooT+30 = 0.7906 T + 79.06
T = 234-288 0C
hfg-!r)\0.09./
9 "=*"7't x 1O1a
R=,J,*" t,, lQ:gt
\ 3.281ff1
R = 694.05 MO
@ ^=*^?2 x 1oi' I0.35l
2r(100cln) \0.25,R=107.'102x1Oen
Solutloru to Test I 19
L)C Electi. Cba.itt 2t
E Not : spdifi c5irt neofth. cabl€ a'rd l.ngth are @nrt?rt iD both conditiors
DCthffikn ==!1nL
&=sR" nC
-,--.lH] .-[#,J
R)=757o
0[r8utOhm'. Lrw alates lhct lhe cunenl florving ln 6n eloctric circuil ls dlr.ctyiilJrtr"ii ii tt'L imprersed enrf applied lo the circuit snd lnv'r'6lv lo lhe
5iui".r.t t .".i.r.n"" & lh€ said orqiil ilamed afier th€ 6€rman physrqst Gcorg
l- Ohn {1787 - 1854).
^1 1
R12G = 0.0833 mho WWmm
ri\6ro E = impr€3s€d voltdge (voft)
I = currenl drswn (ampere)R = r*iEtsnce (ohm)
lltgnltll?affnElectrlcal powor - rale of Using or con6uining the olectrical onergy
Watt - un ol electdcal energy equal to one loula of enctgy consumed in one
iecona. ltarneU aner Oe Bri-iish engineer and inventor Jam€ Watt (1 736 -1819).
,s$uewhere: P = electrical Power ($,att)
E = vohage (volt)I = dfient (ampergR = re3istance (ohm)
DC Ele.tic Cn.uits 23
IUG| C r ltHI IXtXCy toulllErls{. Energy- the capacily to do work
Note 1,lay = 24 boql5l Donth = 30,lays1 yerr = 365 day5 = 8760 hou6
snl$_G0tIIGII0 trsFTtBS.i Series circuit- the resislances are connecled end to end.
+Et- +E2-
tlltlttm$unloi tulD It[ tlsltflns G0rrts Ilr8I t8
E] E.
where W = etectncatene.gy (oute)Q : heal energy (calone)P = etectflcatpower (waitt: time (second).
m = mass ofthe materiat (gram)c = sp€cific heat of the matenat (catone per gram .C)dr = change in temperature of the mate at (.C)
Note l calode:4186 ioales1 BII z 252 .alo:jes
.l Kilowat-hour (kw-hr) _ unil in which etectricat energy is sotd to a customer.
N.n. llth.i..rcthree or more reenances In serier, rcduce 6l'tthecrrc!ii 'nto
twol.ihl,r! lD rn.t b.fore applying the VDr
?rillu+onlcTm ftsBr0ns
a Prnllal clrcult - th€ resistancos are connecled across each olher.
N,n. lniependen{ oF th. .ircuit .oDn€.lion either serl::, parallel or corDbination of1,,
'l h. t hc tota power drawn by tbe circu it i5 equiE lent to the powels d'awn by each lord
l"
,tI3
L
t +E3
DC EleciicCircuits 25
cufin r Dilt$t3r llfittrl IGm ron ffstsnis E0xxrsrrl tr?illl1fl
\o(e .rth.,"a'e.hreF or no,e.ern4nce,.n para,lel, reduce t-l)r (he -tr.u r ntot*o,e< ror. rn pdallFlbeforc applyinq tre lDT
sHllt-pttflfit G0tIIGIMISI$I0[8t Series-parallel circuit-a conbinationat circuit which when simpl iedwj resutt
into a series circuat.
?m|lut$Inl[$ c0xrlcltD [IStSr0ns
Parallel-series circuit- a combinational circuil which when simptifjed wiltresuttinto a parallel circuil.
0[trt r wYt-G0rltcllll Rlsl ols
lE,
U Wya to delta trenlformation3
U D.lli to wye transformationr
u tf A=B=c=Rrandx=Y=z=Ry
_gc- -A.+ BtC
B.c_D.
B,cD.
B,c.D
5cl4Q3cl20
0.5 ohm1.0 ohm0.2 ohm0.3 ohm
Lc.o
26 l00l Solyed Prcblens in Etecbicat
Problc|[ ttr EE Bo.6d Oetobcs r99t. A load of 10 ohms was connected to a 12_voll battery. The curGnt drawn1.18 amp€res. What is the intemat resistance of the batei?
0.35 ohm0.20 ohin0.25 ohmO 30 ohm
Problc|n trrThe potentiat at lhe terminats of lhe banery fa s from 9vo[s when a resistor ol 10 ohms is connecled across s
rnternal resislance ot the banery?
T?tt 2 27
trotl.d ttr EE f,o.rd Apdl a98aA lrd'trrUay power stalion supplies 60 kW to a load over 2,500 fi, 100 mm'?, rwo
rNnrlrr,brr coppor losder lhe resistance ofwhich is 0.078 ohm per 1000 tt The buslfl vollr{,o rB maintained constanl at 600 V. Delermine lhe load curenl
t08 Alt0A
ttalLm 16r 120 V DC motor draws cuffent of 100 A and is located 1000 fl from the supply
to0r|.l ll lh6 diameler of lhe copper transmission line is 0.45 inch whal must be lhell,1146 ol lhe 6upply?
125 32 V130 24 Vl2g 32 V127 05 V
tralLd l7r EE Bo.rd Ostobc. r9a6A LRT csr, 5 km distance from the Tayuman, takes 100 A over a 100 mrn hard
alrawn copp€r trolley wile having a €sistance of 0.270 ohm per ftm. The rail andgtound rolum has a €sistance of 0,06 ohm per km. lf the slation voltage is 750 V,ihrl l. tho vollage of the cal?
685 V690 V685 V690 V
trobl.n larTh6 hol resistance of an incandescenl lamp is 10 ohms and the rated voltage is
00 V Find the series resislance required to operate the lamp lrom an 80 V supply.
V on open circuit to 6terminals. What is lhe
Ftilr
Icl)
llcD
B
c.o
I
ProbL|[ t r EE Board Aprit tt9?_...^^t!"_-"l11rotng!'1"_lT9e ora standard ce is measured with a potenliometer thatgMes a readrng oI1.3562 V When a I 0 megaohm r€"i"t", i"
"onir*r.a ""i""" it"slandard ce termrnals, the potentiomete. ;ding arop" to t.S56O V. Wh;ii"- "inlernal resisiance ofthe standard cel?
174.5 ohms145.7 ohms147.5 ohms157.4 ohms
ProbleE !,4t. .
A battery is formed of fve ce s Joined in series When the e{erna| resrstance rs4 ohms. tne cunenl is I 5 A and wh;n the enernar ,"r,"t"n"u i" S o1.", ;" lrii"nitalls to 0. 75 A. F rnd the intemat ressrance of each ce .
Problca t9rA rcsislive coil draws 2 A at 110 V aier operating for a long time. lf lhe
lcmp€ralure rise is 55 'C above the ambient lemperature of 20 "C, calculale lheoxlemal rcsislance which mustbe initially connected in series with thecoillo limitthecurrenl to 2 A. The lempeBture coeflicienl of the materialof the coil {s 0 0043 per'Cil20'C.
8()4O6C)10 C)
10.12 ohms10-52 ohms
20 l00l Solyed Prcblens in Electical Tesr 2 29
C. 11.45 ohmsD. 12.05 ohms
Ptoblcltr 4OtA carbon resislor dissipates 60 W of power from a 12d V source et 20.C H
much power will be dissipal6d in th6 resi3tor al 120.C jf connectsd across thesource? Assume th6 temperalure coefficionl of carbon at 20 'C is -O,0OO5 per .C,
A. 61.50 WB- 65.21 Wc 6234WD.63.16W
lrroblcln 4rt EE lo.rd Aprlt rgatt EE,Eo.rd Octobcr rt34Two (2) 11$V incandesc€nl lamps A and B are conneclod in sedes acro6s
23GV sourco. lf lampA b mt6d 75 watb and lamp B is rated SOwa s, det€rminethcurlent drawn bylhe sedes conneclion.
1225 m1250 m1240 ln1210 m
tr.t!.o a4r EE to.rd Aprll 1997ll a rcrlrlor mtod al 5 welts and 6 volts are connected across a battery with an
cran ckcult voltago of I volts. What 13 th6 inlemal rosistance of the baltery if lheraaulllng cuflcnl i3 0.6 A?
A 0 30 ohml020ohm0 0.23 ohm0 0 03 ohm
tratldn 4trA hlgh vollage OC tran6mission lin6 deliverc 1000 MW al 500 kV to 6n
{er.Orta load ov6r a dblance of 900 km. Determine lhe voltago at the sending 6nd.Aaauma lhc loop reristanco of th€ lino to b6 1 mO p€r km,
610 5 tV!06 2 kv607.7 kV603.8 kV
tr.tlin '16r
EE lo.rd Aprll lt3t
4.0.52AB.0.64Ac. 0.48 Ao.0.57A
Prcb|...lo 4CtAn arc lamp takes 10 A al 50 volts. A resistanco R is to b€ ptace in serios so
lhat tho lamp may bum correclly from a 110 V suppty. Find lhe power wasted in thisresislor.
A. 800 w'fliB.600wattsC.700wattsD. 900 wstt6
trctLo 4ttConduclor'x" ofs cedain materialand a givon crcss seclion has a resistanc€ of
0.1 ohm p€l met€r and a lomp€rature co6fici6nl of O.OO5 p6r .C. Conduclor ,y" ofanothor maledal and a given cross soclion has a resistance of 0,5 ohms p€r nioterand e t6mp€ralure coemcient of 0.001 pd 'C. ls desired to make s coil havno arc5i6lance of 50O ohm6 and a tompe€Ure coeficiont of O OO2 by u3ing suttait.longths of thc lwo wires conn€cted in geries. CalqJlato lhe requhed length of wiro 'l'
AI0.D.
Lc.D.
ll lr rcquh€d lhat a loeding of 3 kW bc mainlain6d ln e h€eling al€menl at anlntlal lanpareturo of 20'C, a voltago of 220 V b nao€.3gry for he puryo.e. Afigr thoaLmani fu! 3.ttbd down to rloady atato, ll Ir iound lhrl a voLta of 240 voll6 iat!.o...rry to malnlain tho 3 kW loading, Tha olamont reildinca l€np€ralu€o. dcnl l. 0.0008 p.r do9r.o c.itigmd! at 20'C, Caloulato th. tn l l€mp€ralurool lh. h.allng.l6m6nl.
A 345.43 'C4 328.12',Cc 33r| 84 'Co. 318.48'C
taol!.r atrA 200'W, 11GV lncandaacont lrmp haa r flhmonl hay{no r biprnlu.r
@afidcnt of ruBlrllnca aqual lo 0.006 al 0 'C. lf tha nollnal gprr-te ad,l9araturcol tha bulb ii 2600 'C, how mloh cuarlnt rvrll lhr bulb drrw .t lha hda.( i ta lllliodon. Arsume N rcom tcnrparalurc ot 20 'C.
A. 29.42 A8. 18.37 Ac. 22.31,'o. 24.214
hl-ra.tTh. po$qr d..*nbtrr*-a--g-alt tOV..d
10 'C. Calcuht! h ll'e G ly t cj .l 1tt V- 0 'C. Ih.lamparetuE coeilnc,i!.i oli-Fd rO 'C b O.OrOl.
tJ0 l00l Solred Prcblems in Eled,ical
c. 183 WD. 225 W
Probleltl 49!
45 ohms17-2 ohms0.22 ohm4.62 ohms
B.c.D
,-t!F*
An eleclric wals healer has a rating of 1 kW.230 V lhe coit used asheating elenent is 10 m long and has a resistivity of 1.724 x 1O$ ohm-cm. Detemithe required dianeter of lhe Mre in mils?
A. 2.43 milsB. 2.52 milsC 3 21 milsD. 1.35 mils
ProblqN! 5orWhen two reslstors A and B are connected in series, the lotal resistance is
ohms. When connected in parallel, the totatrcsistance is I ohms_ What jslhe ratiolhe resislance Rato resislance Rb. Assume Ra < Rb.
A 0.5B. 0.4c. 0.8D. 0.6
Problqn tt: EE lo.rd X.rch r99a
thsr 2 3I
Nona ol lh6ie
trobLd J4rIhroc rssistors of 10, 12 and "/ ohms, tespectiv€ly are connected in pardLl
.iro.r a constant current source of8 A Determine "x'ifthis resislor draws 2.5A
10 ()12 t)13 Q11o
H
c(.)
t(lt,
scD
trobl.E 55tfwo rosislors A and B made of differenl maleriali have tefiperature coefficignla
ol rol6tance at 20 'C of 0.004 and 0.006 respectively. When connected acrola evollrg6 source al 20 'C, lhey draw cunent equallY What percenlage of th€ lol.lo rranl al I 00 'C does r€sistor A c€rry?
Ihree resislors of 10, 15 and 20 ohms each are connected in para et. What islhe equivalent resislance?
P.oblqn! 5lThe equivalent resisiance of three resistors A, B and C connecled in parallel is
1.714 ohms. lf A is twice of B and C is hatf as muctr as I, find the equivatentresisiance when lhe three of them are connected in se es.
A. 17.5 ohmsB.21.0ohmsC.24.5ohmsD. 28-0 ohms
Problcn 3!Three resislors of 10, 12 and 15 ohms are connected in parallel. Evaluate the
value of curent lo the parattel system thal will make the cufient in the 10 ohmresislor €qual to 2 A.
a7.14%62 a6%a1 34%30.86%
tlcDl.a 55tTwo re6blors A afld B made of different maledals have temperature co€fficiontl
ol rcrislance of 0,003 and 0.005, resPctively. When connected in parallel asosa Ivolhgs source al 15 'C, at consumes equal power, What is the ratio of the pow€l
drawn by rcsistor B to that in resistor A wh€n lemperatute rises to 60 'C? Assurn!aupply vottage is constant.
A 0.829B 0.926c o.s@r0 0.882
Itroblcln ttr EE Eo.rd Lrsft t9teThree resistoB of 10, 15 and 20 ohrns each are conflecled in parallel. whal ll
lhe lolal conductance?
0.217 mho3.41 mhos4.62 mhosO 52 mho
BcD
B.c.D,
12 . !t)!!,1!!!!!t'tohtens ih Eteuticat En
Plobtcor jAr
Problqn 59!
Test 2 33
,3ij,,-i;1,.;1-$i$:tHiyn$:}"?,.,"ki jrf::":*:#:q{sfi
AIct,
s00
615.32 W506.58 W582.45 W604.38 W
llJ ohrn.
B.c.D
tttll.l. atrlrln lho ctrcull as shown, determine the rosislance between terminals a & b.
I(rcont e23 n,oE302362(l
:"9nru;'"m":i*"rl""il";git i';iii,:i'""':"Jii:g;U:;:*i",ilA. 8.10 r)B. 8.52 rlc. 7.84 nD. 9.22 rl
Ptobtqt| 60l
,H:"::#t{tlifl iirstrni{{.{#ifi $;ffi :"ji{lxr#trJ150 ohms100 ohms200 ohrns180 ohm!
h..ll.r arr EE lcad Ostotrr roor
;* ;f,:Hf f ""*:J*;iijttfi"*.xfl{*".,"r,#."J"1:iTfii
j:i:ffi "" *t":iil
a 3t33iiltC. 1a.i ohm6D. 2'1.2 ohmi
Pr.ll- ht ff a-.a l-!t &_
tr "#*ffi{",:;5'; "*i"".sf:t*;I ffi trro,ifi :i?j"f:
;Lu.d C4t EE Xro'.rd Octob.r t99?A 10.ohm and s 2Gohm r€sistanc€ ar6 connected in paraltet Anothern.r.r.ncc or s-ohm is cDnnecrod in 6efi6s w h lhe t"". rr r," *jprrl"ir""" iJ'iivotta, what is ths current through th€ 1O_ohm re3i6lanca?
A I21 Aa 271Ac a02A0 672A
t otl.n| 6titA 30 ohm reststor is @nn6cted ln psre €l wilh e variabte resistanc€ R.
.lh6prr.tlot com_brnation is th6n connocted in s6n€swith
" a"r,Inl."ii.r"i i"o i_..iiirc.o..a l20vsource Find the minimum veru€ of itf ih; il;;il;;';;;l;'fii,"":lo lho pow6r lak6n by th66-ohm rcslslor
A 10.35 rlB 12.24 Ac 10.21Ao 1146(r
Pr.oblco Car EE lc.fit lrlch rrra_, r wo r€srgtanca3 of i0 and 15 ohms, sach respoctivoty 616 conn6cted rn pare glIhs rwo are then connrded in s6ri€s with a s_ohm reeisianc€. tt le than coi;.e-;Jaicross a 12-V ba ary, rhel are li€ crlrr6nl and ,0116r?
acr)
1.2 A,17.28W096A. t152W1.00 A 13.1 W1 5 A. 20.2s w
JP.obLn| 6t!
Probt€lf 6ar
B.c.D
B.c.D
B.c.D.
B.c.D.
B.
5.5 kO5.0 ko6.0 ko6.5 kO
5.32 A5.05A5_21 A5.48 A
4ka3ko5ko6ka
79.1%75.30k
J4 l00l Soltcd Ptoblems in Elechicdl Test 2 Jt
resistors, R and 24 ohms. Determine R if the power consumed by the paconnected resistors js equalto the power consumed by the 8-ohm resistor.
10 ohms16 ohms12 ohms20 ohms
An 8 ohm resislor is connected in series with a parallel combinatior of
A multi'tap resistor R is connecled aqoss a 220-V supply. A voltmeler
tlallar ?rr Ef, ao.rd octoDct t997A procc.! 6quipm6nl contains 100 gallons of waler at 25 'C. lt i3 rsquired to
Llrt0 ll lo bolllng in 10 minules. The heat loss is estimaled to be 5%. what as lhe kWltll e ol lhr hcalofl
t:6 kw,!2 kwt06kwt07 iw
itltat. ttr 3E Eo.!d octobc! lttoA lol.l ot 0.8 kg ofwater at 20 'C as placed in a l"kw elect c ket e. How long a
l[n. lll mlnulol! nesded lo raise the temperature ofthe water to 100'C?
4 a6 filh6 32 mln0Umina 60 mln
tiallitr 74l EE Bo.!d ogtobcr l9glt
1266 kwh1750 kwh1333 kwh
(;lr
Problclr 6l'tA potontial divider of resistance of 50 ohms is connected across a 100 V DC
sourc€, A load resislance of 10 ohms as connected across a tap an the potentialdivider afld the negative teminal of lhe source. lf a current of 4 A flows towards th6load, what is the curent supplied bythe source?
Ploblclr 7orTwo €sistors A and B are connectod in series across a 220 V DC source, Wh6n
a vollm6ter wilh an intemal resistance of 10 kO ohms, is connectod across resistorA, th6 instrument reads 100 V and $hen connected across resislor B, il reads 80volls. Find the resislance of resislor A.
intemal resislance is 15-kO is connecled across the cenler tap and one end ofsupply terminals. lfthe voltmeter regislors 100 V, what is lhe value of resistor R.
Ptobldr 7tr EE lo.td &rll 1992An electric kelde was ma*ed 500 W, 230 V found to take 15 minutes to bring 1
kilogram of water at 1 5 "C to boiling point. Delemin€ the heat effciency of the kettte.
Ho{r meny calories does an electdc h€at€r ot 10o watls generate per second?
l0?3 881000a2 25
tta$.a 75r EB lo.rd oc@b.. t9t7lhc clectric snergy roquired to 6ise the temperature of water in a pool is 1000
lwh It lhr hoat loss€s are 25%, the heating energy rsquned will be
-.
I0o
It:l)
AIc0
I0t)
Problqr ?6! EE 8o.rd April l9'rAn olecldc healer carries 12 A at 110 V, is subfieQed in 22.5 lbs of water for 30
nrlnulss. What will be lhe linal lemp€ralurc of lhe waler if ils initial lemperature ia36',F?
A 135 43 'Flt 125 42'F(; 133 56'Ft) 12a 33 'F
36 l00l Soh'ed Prcblens in Elecbical
lrroblqli 7'l EE ao.rd Octotcr r99oln an electric hoater the inlel temperatur€ is '15'C. Water is tlowing st the rate
300 grams per minute. The voltmeler measuring voltage adoss the heating elereads 120 volts and an ammeter m6asuring cunent tak6n rceds lO ampercs.sleady slate is linally reached, what islhe finalreading ofthe oultel thermometer?
T?-rt 2 37
I lRI
0 !0lRttall.ir att
lhraa r6sislors of Aohm resistance are connected in della lnside the delta
lmlll.l thrac &ohm resistors aae connected in wye Find ils resistance betwe€n any
! ootnorr
2 ohm.I ohm!a ohmr
Four cubic meters of waler as to b€ healed by m6ans of four 1.5 kW,immersion heating elemenls. Assuming the offciency of lhe h6al6r esdetermine lhe time required boiling lhe water ifthe initiat temperature is 2O.C andallfour el€ments are connect€d ifl parallel.
imn€r6ion heating elemenls. Assuming the efficiency of the heater as
A. 57.6 "CB. 68.4'Cc. 72.4', CD. 42.6.C
Probl.ll ?t! EE lo.rd Octob.s ttgt
determine the time required boiling lhe water iflhe initiat temperaturc is 2O"C andthe elemenls are connocled lwo in series in parallelwilh two in series.
A. 275.6 hrsB. 295.3 hrsC.252.2htsD. 264.4 hrs
B. 63 hrsC. 69 hrsD. 66 hls
Ploblcnt 79 EE Bo.rd Octobcr r99rFour cubic meters of waler is to be heated by means of four 1.5 kW, 230,
lI00
Probhltl aor EE 8o.rd Aprit t99?A carcuil consisting ofthree resistors rated: 10 ohms, 15 ohms and 20 ohms
connecled jn delta. Whal would be the resislances ofthe equivalent wyeload?
A. 0.30, 0.23 and 0.15 ohmB. 3-0, 4.0 and 5.0 ohmsC. 3.33, 4.44 and 6.66 ohmsD. 5.77.8.66and 11.55ohms
Probl.rlt ttr EE Bo.rd Octobcr 1994The equivalent wye element of3 equal resistors each equallo R and con
in delta is
31. B 42. B 53. A 64. B 75. D32A43.854.865.D76.433C44.A55.B66.C77.C34C45.D56.867.C78.C35. B 46. C 57. A 68. C 79. A36 B 47. C 58.8 69.8 80. C37A48.C59.A70.C81.C38. C 49. B 60. C 71. A 82. B39. B 50. A 61. A 72. O40D51.D62.A73.441 A52.8 63.8 74.8
qg-5? - Top notcher
AE-35 - Conditional
E=*,=E-s=-11-1eI 1.18
= 0.169ot :0.2O
r=&=9=oerR 10
v =E-vR =9-6 =3V
I 0_6
r=5O
, vR 1.3560' R trlo6l=1.356x10rA
=1.3562-1.3560v = 0.0002 v
v. 0 0002| 1 356x10 "
t = 147.5 A
Lel: r = internal resistance per cell.
,E5r+RL
Substitute condilionsl
15=-l5r+4
E=7.5r+6-O075= l
5r+9E = 3.75r +6.75 +O
EquateEq.ltoEq.2:7.5r +6 = 3.75r + 6.75t = O2r)
Sllutlonrtolts[2
R:t0t)
R=lMO
tt 100) Solved Ptublens in Elect cal SolutiorLt to Test 2 39
R: t00
r ;.1'#::*:i:fi'"-"-'"
I. - ' +lRi
ooo-4!99*rro.gsrI
600l - 80,000 + 0.391'
tr 1538 46+153846= o
L.,.lng lh6 quadratic fomula :
. 1538 461fi538.46r - 4{1X153846),- --- n1538.46 +1323.433
N,,t. Vre negativ€ rigD folminimum curent
I - 1!!!39:.1!39-1!9 = 1s7 513 x2
lq 1O8A
Let R = tolal resistance of the line
A . d, =(0.45x1000)'?= 202,500 cM
R-"!n -]9!21?49) = o roz+ o
202,fioE. = VR + lR = 120+ 100(0.1024)
E. = 130.24volts
Assume iR = equivalent resistance of the circuit
Vn = E' -lRv. = 750 -1oo ft0.27+ o.o6x5tVR =585V
Lr-+--1000 t
E,,= v50R, 10
=E-r:lL=99:!9=eo
I
--------> Ft00A
10 l00l Solved Prcblen: in Elecbical
a.=f =f =so- n,f + o a,)
*Rr=R,
R1 = 44.48 ot+od, 'l + 0.0043(55)
I
rl
R = Rr - Rr = 55 - 44.48
R - 10.52fi
Ar 20€:
n. = l4 = 13{ =zoo.,'P60At1200c:
R, = R1(1+ aa, ) = 2ao ! - o.ooo50 20" - 20.)l= 22s n- E2 12o'
R, 228
P -63.16W
R- = Ei = ll
Solutio,te to Test 2 1I
h
r;;'t'
rl
T
11 +R, . lQg
i, . !00-R, =1p
lqu.[. Eq.I io Eq.2 :
ir(l+0.m64) + Ry(i +o 0ol4) = (Ri +RyXl+0.0o24r)
ir +O,OO6 'R,
+R, + 0.00larRy = R" +0 0024rR. + Ry + 0 002d,Ry
C,O3R, -O.mlR' =0=@
lubdtlt Eq.3in Eq.4 :
0,00!Rt -0.001(500-R,) - 00,OaR' '0 6
ir.126ohme. t26nr. '5;67Lr . t260m
r.f .{=z.zn"P5.E' r+R
,.E-n=9-z.z'| 0.8
r. O,3o
r. (0.001)(2 x 900) - 1.8 o
t-l.1mox10.6 = 2ooo AV, 500 x10'Er . V + k = 500,000+2000(0 18)
E. ' 503 6 kV
R.-:L=:::=18133O' P, 3000
"^=Ei=4=rn2n''' P, 3000
R, - R1[+ o (1, - t1)]
19.2 = 16.133 h + 0.0006 (12 - 2o')l
t2 = 336 84"C
3.t8 occ . ll[(.oz - zr) sooo o+r€o! sr. z 6r
[(,}- tt D+qru - tuoom zd -uzst=TA=-E="8
ome zdUeet Sl=-;;= !:='U
Argeoe-13(8! o)oooz + ooo'ms = rt+ h-'3
.0!x 009 hv0o0z = ^m = d.l
u s.! = (006 x zX!00 0) - I
UC0=I80 IZ/- 9 =U--=r
u+lI "l
9du 11-; = -:;- - u
ur oez! = ''lui/ut 0 ,_fi;zr= |
gurqo g?t- rU
I o = 'utoo oo = ('u-oo9)loo o-'uooo o
:tbf ute blqnuaqns
@=o= ^ulooo-'ucoooru'vzoo
o + rU+ 'u'vzoo o* 'a = ^f 'vi,oo o+ ^u + "Urv900 O+ 'U' ('vzoo o+l)(ru+ 'u) - ('vroo o+ t)ir +('vgoo o +!)'u':Z b3ol!'b3 elenbf
G)- 'u - ooe = 'u009 = ^a+ 'u
It a ,tal o, sltolt vos
a2 1001 Solved tuoblens in Elect ical
E *,=f,=ig=*' '1
'Tr=- L=zoo.c
o.oo5
R, _ T+t,R1 T+tj
*, = *, ffi ) = *. (r...,.*_ry-) = .,. "
,E110R, 4.93
I =22 31A
E *,=#=#=*'
e =ra'1
Eila
n,* = n.! + ca,]= ss[r+0.003s3(100 *20j = 72.29o
- E? 11s'F,oo=n- =zz:e
q* = ieaw
.=f,=ffi=u,nnn=o!
A = e!= 1.724x1:0::(0x100) = 32.sg9x1oacm,
4{32.589 x 106) ^ ^^--- lo0omits6dnx_n 2-S4cmd = 2.52 mils
When aeri€s connecl€d :
Ra +Rb =36+OWhen parallol connecled :
El*;='-o
Solutiotll to T?^tt 2 13
lub.tltut. Eq.I to Eq.2 :
R.(ra-R.)-8(36)i.'.30R.+288-o
UdnC thc qusdretic fo.mula :
- ae r '6rlfoxzs8) 36ii2 36-12
Rr.36-12*24o
nrlo - l!21
Rrtlo - 0.5
1111Rt Rl R2 R3
ttl1Rr 1o 15 20
Rr-4615o
1111R, R. Rb Ro
11111714 2Rb Rb 0.5Rb
R! =6OR. =12oR"=3o
n,t Rr* Rr
tqa t 15r,1 200n,
wh6n connoct€d in series :
Rr=R"+Rb+R.=12+6+3Rr = 21o
I n=ffi=ffi=uu'nByCDT:
, I,R
"=R,+E, | (R, +R) 2(10+6.67)\= R = 6's7
\ =5A
12 1001 Solvcd Problens in Ele.tical
n, =4=]l{=*..,'P2001
-TT=---1-=2oo.c
0.m5
R, _ T+t,Rj T+ir
*, = *, (Jf!.u J
- * u(ffifr) =.'.'.E110
R, 4.93
l=231A
'o=Io"
'=',F=d= 2.52 mils
When sedes connected :
R. +Rb = 3s+OWh€n paral[6]@nnected :
El*='-o
E *,"=g=g=*''n,* = n.! + ca,l= ss[r +o.ooseqroo -eo] = zz.ze o... E2 1,1.s2n.=C- =rrfiP,- =ioew
q = E1= 3g!i = sz e.,P 10m
n =.!A = !! =
1.724 x 1:0-6(0 x 109) = 32.s89 x 1o€cm2
4(32.589r10€) - ^---- 10O0 mits
r 2.54crn
Fr'Solutiora to Te$ 2 13
trb llul.Eq.1toEq.2:i.(16-R.)-q36)i.'|-36R.+288=o
UtlnC lhr quadrallc formula :
t.- o i .,..6t- a6es8)2(1)
Rr.36-12*24o
rrlo - 1121
R.tb.0.5
't111Rr Rr R2 R3
1111R, 10 15 20
Rr .4.615O
1111R, R. Rb R.
1111't.714 2Rb Rb 0.5Rb
R! =8OR. = t2oR. =3O
Wh€n @nnecl€d in s€ri6s :
Rr =R"+Rr+Ro =12+6+3Rr = 21o
n= R,R, = i3M = o.oz oR,+Rr 12+15
By CDT I
, I,R
" = R;;E
. lj (R, +R) 2(10+6.67)
"= R =-- a-?\ =5A
36r12 3a-12 _-
22
& n,{ Rr{ R,t00t tsat 200
E,
aa I00l ! ued ptublens in Etect cat
n."ffi=1wa=s.*nByCOT:
, l,R''-R,-R
=f-n=!.(!!9 5a5
Solutions to Test 2 15
=12()
l. = 52-86% of lr
At 100.c:
At 100.c:
=Re!+o.ooa(loo 2ot=1.32Rla
= R1b11+ 0.006(1oo - 2o)l= 1.a8R$
-----"> I, -2.5 A
Not<_\hethev.lrawedudr .ur..,rhat20.C. r\en r.,= p,!, pAyCDT
R,.. L tl.4aRlr. = -;R;-;R =0,5286rr
R.
l{
- E-'- = 1991 = 13.15 eP, 760
_E1=lq=658,,Pr '1520
,, E' 1oo2'"' R-- - 1316+6.58|k.. 506.58W
ll - Pl +P, +P. = 750+250+200P, - 1200 W
r. - Pr=f?99=ror
' r't 12O
1,-Pr=I99=75yr, 10
,t,lr,
x--E1=!{=zoo.,'P,50
n"-E1=194=roon' Py 100
N,n. !J thrt th. hmp5 may opeiate prcpefy when connecte,l in se;ie5 too5s a,'(){) V ntuilc, thc votage a*oss each amp must be equalto lOO V. rlus, the,Arn,rL.acos: thc paralcL bEnch irust bc equalto thc rejijtanc. oFlamp y.
u -L
I', 2P, +Olt, t Pt = 2280)O
lirb8tilule Eq.1 in Eq.2 :
l', | 2P, = 22AO
t'. 760 wl" 2(760) = 1520 W
- 120 75 = 45V
-E; - 4s'P, 25O
-81cr
Rr. -R1.(1-oa^r) - Rr! * 0.003(60 - 1st - t.135R,"R2b ' Rlb(l * ob^r) - Rrlr , o.oosloo tstj. r zzsn,o
Note, Since they daw equal powe6 ar j5 .C. rheD Rr= &b_ R
o -11. . E':
P. R,, 1.i3sRE=F;="t* =on'u
G- =1'Rl'l l1l
R, R1 R' R3
^11110 15 20Gr =O.2167mho
R,l R, NJ
200tqal rsn
l" nJ
aa lql Solwd Pmbf.nt tn Etectricdt Ekoi
1oo. !.eglR+20()
R+20O=2RR=200o
E'-#*i.*R' -678o
toa
10a
80
^,'#**uRr -11 o
R." = 1+ 2(3X4)- 2(3) + 3(a) + a(2)
Rn = 1.e23o
tSolurior,.t to Tert 2 17
nr =ffi+s=rr.ezo1. -Er=-IL-rrrae' R, 11.67
By CDT :
, f, R, 4.113t2O\
"-R,*&= 10-m\,2.74 A
s"cor r" =]i4l- ' 3O+R
trr(6) = r.'(R)
'*--(#ffi)(ao + n)'! = T900+00R+R'?=1a,oR
Rt -goR+9oo - o
!y quadr.tc fumula :
- co t Ji6F - a{teoo)"._ln
EI48t/
i
Ii
T
g0 i 87.08
Tala G) for minimum value ofR:- 00-870a
2R.lt.46 n
R,..lgM*5=11o' 10+ 15
r,.f,L-lf =r.oor
Pr . Eil' .l2C.0s)
A .13 08w
E,t2v
1,, ,2
lE I00l Solved prcbtens in Etectticat
_ t.R' R+R,
r, = r.(Rf R.)
= qnJ to)
= O
v = Er -Vr = Vr -|LRLV = 100-a(10) =60V
,'=#=effi-aEquale Eq.land Eq.2:4(R+10) _ 60
R sO_R(R + 10)(50- R) = 15R
50R,R, +500 1OR = 15R
R, -25R_500 =o
Solutions to Test 2 19
\ote 5rn." power. d%wF are -qua' rhc equralent ..s.<tan.c.omb,natron mlsr he eqlalto I ohms:lso
R"q=8
8=l1L24 +R
24R = 192 +8RR=12rr
lj = l, +1,
120_ 100- 100R/2 R/2 15.000240_2OO- 100R R 15.OOO
40 _ 100
R 15,000
n=ao1#99=oeoo
R=6ko
by CDT:
U.lng lh. quodretlc lormub :
zr , Jzsi,lttx sool zs*st.za., 2(1) 2
26 r 61.23'2R - 38.12 A
&rbtlllulc R inEq.2:80 60
'' 50 R 50- 38.12
t, - 5.05 A
t20I/ t00 y
Urlng lho lirst condition :
100 1m 120
Ra-1qooo= Ro
1- 100, + 100
Rb 10,000(120) 120R.
i; ="it.";=oUring the second condiLon:
t#=*m.#=u
substitute Eq.l inEq.2:
r4o so ^^l t s)R, =
r oooo . "'[ r rpoo
- 6R" ]
140 1t 400
R" = ?50.6R"t14o tt mol^l*" =t*-u*"J*140 = 0.0'146677R. + 66.6667
R" = 4999.88 O
R.:5kO
---'>L """"> It
Er =20 v
h =220 v
E o = 0.24Pr =0.24Ett
J.
Q = 0.24(1 10X12X30 min), !9!99 = szo.zr r""r
E Note,l atore = 4.j86./
n- Pt loo{1)- - 4Js6 =;.i86Q = 23.98 calories
Wn = 10oo+o.25wn
Wn = 1333.333kW-hr
!! ! AJ!!!!!!!!!!2!!r, i, Et""t,i""t
Solutions to Test 2 SlO = (lOoO g)(4 186 J/g ,C)( 1Oo _ 1 5) = 356.810 .t
,r =l5g!=- 3558r0w"* 5oof;;*]=o?eo6\=79.1o/o
. r00s"r,,ffi"ffi,ff_a,are
X:- il1'.itj: !T.T[',?']lo. 25) - 283s0 kca,
cl,n = 29842. 105 kc€t
o = o.24 Pt
zesrz. t os = o.zap f1o ,nin " !9lgg l
P-2o7.24kw t tmin J
Q = mcAr
Q= (80osX4.lss J/g.CX100 - 20) = 2679o4 J
,w2674' - E = -i6o- =,ut n*s.c x =!!LI = 4.46hjn ou sec
fir - .J/ orbt 2
1h = 10.22kS
r, - !1r :zy- f,1:s- szy = r.mz"c
l] -, r: r lnc(1, tr)| () 570.24I 0(, I, =
10-.(1)+1.667=5746.C
L - f'c,::= e(slrot *.,
l, - tJ5.43.F
{J - n,c^, rO(), 0 24pr = 0.24Etr+O
Ltriats Eq 1lo Eq.2l0 24Etr = mcA,
o zler= f("a,)
o zo1'zoy1rel = p9!,-1rL
111g,r l,^ .\ min 60 sec J I s ,c ./,-'
,
A 57 6oC
l, = ar -!1 = 57.8_15
\ =72.6"C
Pd,M = 4(1.5) = 6kWP*a* = Pon
o I = 6(0.9) = 5 4kWO = mcdl +Oo=Pt+o
'n = 4 m.rlqg& l= 4ooo i"
equate eq.t roeq.Z:mci! = Pt
r- mcA, _ 4000 ks {4.1s6l{IqI).(!g:!qlg
i = 248059.259 sec
E Note, rle iating ofrh. hearo is rlways the power drawn from rhe suppty
32 l00l Solyed Prcblens in Electrical Sol tio8 to Test 2 53
I = 248059.259 x thr = 68 9h.g
3600 secl:69 hrs
R =; = tsoo = 35.266 o
t dre 'atrng ofrne heare. rs rlway,tlr powcr,l.awn 6or the suppli2 Vlheh two Scaterecncnt. are.on.ected In € e!. {he'ort,:9.a ros. eacr
element wll be equalto l,/2 oFrhetotalvo hse2 5,1c. rct spe.l'icq. a*Lhe (he. rLir -onnF.l,ol qoe. no, a4a trc
e#iciency o(the heato.
""* =.[+] =.[-q*] =,uoo * =, .**P6ftdtu = PdM n = 1.5(0,9) = 1.35 kWQ = mc^l +Oo=Pt+O
. = + ," fl!99-9) = rooo rot 1m' ./
EquateEq.l to eq.2:
t_ mc^r _ 4000 kg (4.186 kJ/kg oCX100 -20)'CP 1.35kW
t =992237.037secx thi3600sec
t = 275.6 hrs
e- 1o(2ol--lllc,t0 + '16 + 20
r. Jo(16) .rsso- lO+16+20
c- l6(?Q- = s.eeo
lO , 16 +20
lr-rR R,
i,-:
Convcn lhe Y resislors lo a :
R^ - 3Ry = 3(6) = 18O
p - 1!{9I=a5e' 18+6
Ry - 4.5+ 4.5 = 9O
,. R,R, 4.5(9)"* 'R. -& = 45-,Rrt =3rt
Ir
-----------xF-
51 1001 Solved Problens m Electrccl Netvatk Ldvs & thetens SS
',1,,1 1 , i...lr!l lr.ric cvr uate.lby tDultarco!t trbstjt!irons ofthe, I' rt,r r l,rtrrLrl.rtc.l,A n9 KCLand KVL
lllxwltls lsll flIl0[lhll
'r.lhod lnvolves a set of independent loop curent3 assagned to as many
itath .r rl oxisl in lhe circuil and lhese curents are employed in connection wilhlftxot rrln rosislances when the KVL equations ale writtentrncitoF$ uw
Named afrer the Geman physi;$, cusrav Rob€.t Kirchhotf(1g24- 1887)
,, CunEnt L.w (KCL) - the atgebraic sum ofthe cur.ents al any junction orof an €lectric circuit is zero.
t Voltags Layv (KVL) - the atgebraic sum of th€ emfs and the resistancedrops in any closed loop of an electdc ciroit is zero.
Sign Convenlions for Kirchhoffs Law:
O Currenl towards lhe node, positivecunsnt.o Cufi€nt away frcm the node, negative curent.g ln a vollage source, ifloop enters on mrnus and goes oul on plus, positiveCl ln a voltage source, if loop enters on plus and goes out on miirus, negativeO ln a resislance, if the loop direclion is the same as the cur€nl direction.
n€getlve rrabtance volta06 dropO ln a resistance, if the loop directioh is opposite to the clrent direction,
reriatance voltage drop_
n/
6^r,R)
@Itlvl
, ''t'A Ei - lA(Rr+ R,) - lBRl = o,. 't'lJ l::- lr(Ri + R6) - lAk = O
M.\h.!rent5la and Lua'ee€luated by 5im u ltaneou, 5u brtitutions oFthc
-1 ,.rt ons lotmulate,l6om each oopormerh using KVL
q;lqi q"]rl
s|PtlP0sm0x [oRfii
h r r$lwofk of resistors, the currenl in any resistot is equal to lhe algebraic aum otlha curronts delivered by each independent sources assuming that each source is
tolhr0 alone or independently wilh respect lo lhe olhers.By KCL
By KVL :
E1-liRr-lrRr=OE: - trR.1, lsRs = OE1- lRj + l,&- Er= O
56 I00l Solved Prcblens in Electti@l Neteork Ltus & Theorens 57
sTEP 2: lfE, i5 acting alone,
Note , ltre cunentr ll', l:', L', l1', l:' & 13'arc cqluate,i ushg basic electrc circd(
tottlrou flllo!lhl. m.lhod, r clrcult wth "n" nodo!, has a solulion with only
lill.llon. nacd.d.
Note, lFa 5ource (eithda cutrentoia voltagesourcc) rs acting alone, ihe othercureDt sources a'e open circujted whileth€ other voltage sources aie rhod
5TEP 1 : lfEr ii acting alon. Connon node
\rr. tl)crc aicthre (n = 3) node5 rn thrs crrcuit, therefore on ly two equitions.'r. r.al.d toiolvcthr5 PDblem
| {l i!rl. b, lr = l! + 15
Nd,lc voltage5 v.and Vbare daluated by sim ultaneous substitutiont oFthe
.quitioni f;'mualtcd using KCL an,l otespondrn!!, currctts flowiDg througb
tr.h rcrstancescan be solvcd.
If,Iwilt$flI0nilll a taalalor of R ohms be connected between any lwo terminals of a linear network
ll. ra.ulling steady state current through the resi6lor is the ratio of the potential
allLr.nc. E. (betw€€n lhe two points prior to the conn€ction) and the sum of thoYalu.. ol lhe resistance R. (resistance b€tween lh6lwo points) and lhe connect€dlltla{anco R. Named affer the French telegraph engineer, Chatlea Leon Thevenin(1667 ' 1926)
L
I
St l00l Solled Ptublens in Electical
Thevenin s equivalent circuil:
Example, solve for tr.
0
sTEP 1 | Open circuit &, and:olve for the voltage acbsr the op.n crrcu(ed t€lminals.
Note, Eo rs computcd uring any rhetbo,l5 (Ktrchhoffs, Ma&e I, Nodal, etc..) ofaDalyziog oetwo* Problcmt.
Netwotk LNs & 1'heorcns 59
Conprin t (rhort circutallhll.Fn{.nt curcDt tourc.t).
wlahcornputcd using the basic priDciple5of
llwn clrcu(.
(on u.l thc Thev.nin': equiv?lent ciEuit and solvethe resultiog cunent
RO R3
rcnmffilf,t0ltxl! analogous to Thevenin's theorem except thet instead of lhe open
, lhla theorgm uses the Bhort drcuil te6t and tho 6quivalent circuit is aCmllt. Nomed afrer lhe Amedcan engineer, E. L. Norton (1898 - )
aqulv6l€nl circuil:
iodcpcndcnt voltage source5 and open clrcuit all
iindinq thc tota I rcrista nce ofa
t,
RJ R,+
h
l"I
50 lo0l Sobed Pmhlens in ElecEical
sTEp 1, short crrcuii Rj ?n,1 soh€ tbe sholt .irc! it currcnt that llowstowardsthe
N.t\|o Laws & Theot.ns 61
ruxl $fltaltrnumbat ol voliag6 aourc€s of arbrtrary geneBled vohage and finhe
iiinoo ddhrcnt trim zero are connecl€d in parallel the resulting voltage'.'dh__-'_toomutnatton
ia ttre ralio of lhe algebraic sum of lhe currenls that
Siidirioir. v 0.r,""r. "ten shorl circuited to the algobratc sum of the
lIltlolanc.
tQulvll.nl clrcuil
E,
Note l'.,5 computed using?nymethods (Kirchhoffs, Ma"well, Nodal, et..) oF
analyzng octwork poblems
5TtP2, Compute R.
Not., R.6 computed uting the basic principle, of Fnding thetotalresistaDc. ofagiven ciruit.
5TEp 3, CoDstiuctthe Norton's equi!4lent circ! it and rolve For I'
l^,
62 l00l Solved Prcblens in Electlical Enpineehns br R. R
5TEP 1 Drawthe Millman sequN?lentcirco( and solve for Vrb
5TTP2 5olve for lJ. b and l:
stuncr rmrs[0[ttll0I ilHIt0!Thls method srmplifies lhe numb€r ofmeshes rn lhe nelwork and thus simpliliesthe number ofequations needed.
* Voltags source to curentsource
tre
., Cunent source lo voltage source
Es
Rs
It./l,'or[ LaN &Thzon s 63
tblw (or ll
thc rcltagc and cuncnt sourcet to th.lr cqu i€lcnts'
ltmplltthc Ptallel branch andtransform bacl<th' cuiient sourc'to an
qutvrlcnt voltage sourcc
--------> It
llP ! By K\iL (utng thc aivcD looP d irc€tioD ), tt crn thcn bc solv'd
9&-1"(p. 'p" " p.)- 1"p. -s
61 I 001 Solved Ptubtens in Electrical
xt ru t0xftfiIr $Ennr0nlMaximurn power transferred to 6load rss,stor occurs ontywhen the sad resistor1-"j.tu.,_"q*t to the resistance (R.) of the networi ro"t,ng u""k f._-t;"'
Test 3 65
Ex?mple Find R iFit 6to absotbe,l maximum pryer.
solltion: o?enrrcu( rcsistor R, shortcrcqit ail independeDt volhge, 5ource5 aDda lI in,lepen,lent .uneDr sou rces.
Rt RJ R,
.Ro
No(e, ifrhe @aesuo'rl "s .Lr.c1, or po*e. drah') by,c!s,o, p 5 to be ro .ed ..eelhe the\en|n ror \o,lon s thco,em ro.o.ve tre p.ob.em
aL- alr EE Bo.rd Oc3ot "
r99CA 12 V baflory ol o.os-ohm resFtance and another batteryta.r.r.nc€ supply power to a 2_ohm resistor. What is lhj
llnA103 Atf2 Alt 91 A
of 12 V and 0.075curent lhrough the
04r
A, zrv| 2sv0 25v0 3ov
IcD
p'ip.:itn1["str":,fi3"f,ffi:1"}[,s";j;H:x,lr"J?"",ilffi i
trobls! a5! EE Bocrd At dt 1989Tho LRT lroll€y syslem 10 mites tond ir fed by two substalrcns that generate
100 volls and 560 votrs, rospectivety. The ;esistanc€ or lhe lroley wire and rait ler ntl o 3 ohm per mite tr the car is toiated 4 mite" r.o. U," OOo*ilr "i"iii"-oi*J)"riii
^ lrom the line. How much is the cunent supptied by eactr station? _ - "
133.33 A, 66.67 A123.67 A, ?6.33 A1 17 .44 A, 82.56 A125.54A.63.05A
Problqn 36tTen cells each ol emf 1.5 volls and inlemal resistance of 0,2 ohm are ioined inpamrer and connected to an exremar circuit resisran* o i ir,rn!. #r,"iii,i#iiiili
A 0.45 A8.0.65Ac 0.50AD 0.48 A
R3-
Is
66 lOAt S.)lr!Lr 1,tu1)k"8 u-Etetunol t hgtneeting b R
Two battedes A and A are connected across each other with lerminats ofsame polarity together. The open ci.cuit emf and internal resistance of each b;ijs respeclively 24 V and 1.5 O. Oetemine the resistance ot a healjng toad conn;iacross lhe parallel combinalion of baneies so lhal tne power mnsimeO in tne t
Problem a7:
is 100 watts.
B. 422()c. 4.o2 0D. 4.12o-
Problcm 8a: EE f,oard Octobe! rtx)oThe lead storage bafleies'A" and "8" are connected in para et. .A, has
open circujt voltage of 12 V and an inlernat resistance of 0.2 ohm. Ba(ery,,B,, h;;open c'rcuil voltage of .12.2 V and an internat resistance of 0 3 ohm: tf the ib€ efles together detivers power to a 0 5 ohm power resistor. Negtecting effeclstemperature how much cLirrent is contibuled bi baflery.A"?
Three resistors of 2 ohm resistance arc connecled in della. lnside the d(lTlti ]l':: 2 ohm resisrors a.e connected tn wye. six oatrertes oi neirigiinle.nal resistance and ot different emf are inse.b; i"6 .";;;;; U"Y;;;ammeler. lhe current rn one ot tne defla Dranch was tound oul ro be 3 A. tt a 4_;h
a.c.D
B,c.D.
Plobleln 89!
resislance is inserted into that branch, what wi be the new cunent?
A.2.00AB. 1.75 Ac. 1.50 AD. 1.25 A
Problclr 9Or EE Bo.rd Apltt r99"
train ls al lhe distance of minimum polentiat?
29.62 A16.00 A12.85 A25.24 A
175 A, 125 A183 A, 117 A164A,136A172 4.128 A
_ ln Man a, the LRT runs between Grt puyat Statron and layuman Stallon wh.
::_-.^l(Il-?:l_i"9 Tlllta,ns v^ott.ees of 420 votrs and 410 v;rs respecrivery iresrslance ofgo and retum rs 0.05 ohm per km. The lra," d,"; a;;;;i;;i;ijror ruo A whrre in motion. whal arc the currents supphed by the two statrons rf
Ili0
AIll0
t(:l)
TPtl3 67
tFtl.m 9rl EE Do.rd Octobcr t9a5' '-rr r iii i;llr, s tm oistance fiom lhe Tavuman, lakes 100 A over a 100 mm hard
*rrri ,.'l'r- irottev wire having a €sisl;nce of 0.270 ohm per km The rail and
lii"-r '"i''" n* "
*"i"lance df 0 06 ohm per km lf the siation vollage is 750 V
llnl ri lrr.,'lliciency oltransmrssron /
,r.u.rn 9tr- '-a 'n"'"iun"e
n '"
connected across two batleries A and B connecled in patallel
fm """rii"i,n s.rs ana inlernal resrstances oflhe banenes are 12 V 2 ohms and
i t t "lr" . respeclrvely. Deler mrne the ohmic value of R if lhe powet absor bed by R
l. / lllit wAlls
l) l)ll ()
tr.tl.D ltt EE aoard octob.t t96t- -
A ,,harqer, a battery and a load are connected in patallel. The vollage acrcss lhe,
nlrl0r rs i2 6 vons anO the banery has an eml of 12 volls and rnlemal Iesistance of
ili i""' r"" load consists of a 2 ohms lesrslor Frnd lhe cLrrtenl through lhe
{r (il Ar|2540424o50A
,"obl.D 94! EE lo.rd octobcr lt9a_ lead ;l;rage battery is €ted al 12 volls lf lhe internal lesrslance is 0 01 ohm
rllntls the maxi;um power lhal can be delivered to the load?
l 3.600 w{: / 200 wt) 1800w
Probrer 95! EE B(|.td Attttl a995a l2o-tiaterv havrnq an inlernal resislance of O 5 ohm is connecled lhrough
n h,(, resrslance of9.5 oh;s io a vanable load r€srslor' What maximum power will
lh,, ballery delivet to the load resistof
68 1001 Sotyed pnblens in Etectical
B. 63 wattsC.630waftsD.360watts
P.oblqn 96t
toor EE Bolr'd Aprll t99t
torr EE Boord Ostobcr l99l
lw.lv. .imilar wires each of resistance 2 ohm6 are c.nnecled sollfid lh€ r6sislance between the two com€ls of the sam6 edge.
I ll3 (lI t02 (lM6oI l0l l)
Test 3 69
;]urfr lTi"jJ,l#j"!T.:"T,:;;X?ll:r1x,:"i::iil;til,,":#itr:*l,Sgiffi ,,'3;"'j:, ;-ifffi
"''":"".#'.T1"rJ T B'f5|",f '*:A. 8.0 rra. 1.5 f,rc. 2.0 crD. 1.8 ()
# ?#t?i:r:i,i!l ffi:1""1""1*3""':'i"i v,11,:f ._iffi,i#,,;.,:
0.25 ohm. sorve ro, the m",_,,i'i"*iiiitl,i# ffffJ"yf,i:,9jeli*,
hrlv. idonticalwircs of resistance 6 ohms each are arrang€d to form lhe edge
I trbr A curent of 40 mA is led into the cubs at one corner and out at the olherly opposite comer. Calculate the potential difference d€veloped between
PlolFln 92, EE Bo.rd l.pdt rrsr
;T';[L",!i{fffl.r.':.ffi :i{irri}$;:3l-".", ll,?:*:#A. 230.77 kW8.220.35kWc.242.73kWD. 223.94 kW
Ploblqr g8r
tqr EE Bo.rd Augun t97Ctlnd the value of the voltage V.
0lo v0riv022v02av
L 31Vfi 24V12 19V1166V
-o.215 A0215A-o 306 A0306A
Lanp60wt2v
A. 130.20 wB. 115.52 Wc. 120.21WD. 142j2W
Probte& t9l EE Bo.rd Aprtl r99r^,.h-rweNe
stmjtar wires eech of resislance 2 ohms are connected so as lo formuuue. rrno lhe resislan@ between the two diagona y opposile cornem
21v
rots EE lo.rd Aprll t9ttR.lcrring to the circuit diagram below, if the chaqer voltage is 130 volts and tha
vollage is 120 volts, solve for the cunent Ib.
B,c.D
I
1.45 ohrns1.66 ohms2.01 ohms1.28 ohms
BatreryChatget
0.10
+
70 l0Al Solw.l Ptoblens in Itlecbict
]iF
by R. Rolas Jr. Test 3 7l
Problcm ao4r EE Bo.rd August ag1zln the ligure below Rr = 1 ohm, R2: 1 ohm, R3 = 3 ohms,
ll.i loTliolv. lor I uslnq Sourc€ llansformalion method
V. Fand Es.
A. 182.41VB. 153.32 Vc. 164.67 VD. 157.22y
0trA
03llAR/
AI,00
B,c.D-
211/
Problern 1o5r EE Bo.rd Ostobcr r9ao, EE Board Apdt r9a4ln the dc circuil as shown, the high resislance vottmeter gives a reading
+0.435 volt- What is the value oflhe resistance R?
B. 5 ohmsC. 3 ohmsD. 2 ohms
)-b- roct- lolvo lor V ustnq Maxwells mesh melhod
I?Vl6vl4v
25460A33A
t2v
Ptobl.rn 106! EE to.rd Aprll rgto hrDtoo ro9:'--rr.i.*in" tt
" "un"nt rn the 1-ohm resistor using Norton s theoremDetemine I in the figure-
0.028 A0.010 A0.025 A0.0't4 a
IR?
Rr+
------.F-!2 tl!!.!:\:t::!tn*ren.g
n trec -!u!1ty!!g by R_ggt1,
Ploblcllr ltorDetemine the curenl in the 1o_ohm resLtor us0.8334 ing Thevenin,s theorem.
0.667A0.707 A0.508 A
a.c.D.
t0y
Plobte& lrltOeleffnin€ Ihe;urent in the 6,ohm resislor by Superposition theorem.
B.c.D.
4.2 A4.O A3.8 A
Probledr u2!D€termtne the looking back resislance belween terminals a & b.1lo
B.c.D
9f)10o12o'
30r
3t2
t0!)
Test 3 73
hurnrur
AI00
lirttrm rr.lrl,otormine lhe vottage V
I2v
I2 t'
A 97. A 1O4 C 1.t1.A 98 C 105. C 112.A 99. B 106 D 113D 100. D 107. A 114B 101. A 108. AD 102. A 109. CB 103. A 110. B
l2vl0vt0 v
D 90.c91A 92.c 93.D 94.c 95.c96
BD
c
aA-3P Topnotcher
l,E-el, - (onditionat
0- 1,5 - Failed
74 l00l Sobed Problens in Electical Solutiotts Tat 3 75
Apply KVL on loop'dabed
VL = zl40 volt6
, 600 v, 600- 4401.2 1.2
l, = 133.333 A
t" 560 VL _ 560 - 440' 1.8 1.8r, = 66.667 A
lJse Thevenin s lfroolem:Solve for the looking back resistance acrcss lhe load terminals.
Note opeD circuit R!. 5hort-ciicuit allthelocclls. By in'pection the lookinq back
'^n,ar.e s equa I to 'l.c egur\a lcnt .esistance of 10 rdenriLqIre.ftols -onn;e,l n
n^ = 1= 93 =o.ozn-10 10
Solv6lhe open circuil vollaqe across lho load lerminals:
\'tre Srnce the cejh cr rdentrcal therc s no ctrcrd{rrg cureri in,ouqn thet'trdl e .ohbin"r.on, thui Eo i. equal rorhe ope.r c,rcu,r.mfofore ceit
Eo = 1.5 volts
U8ing the Thev€nin's equivalent circliitl
r, ="|.t=ffi=orsooll :o.sA
N,,ta:5rncethecellsare identjca, the load cuffent 0r) 5 djvided equally 0/1o).r,,,Jr!thetcncel, Refeitotlecjrcuit,{ragianl, conydera loopwith
"necell, ) y r')d ihe loa,l ie5r5taDce.
Sol&BbTlrH
12-O.O5\ -2\ =0I = 2a0 aoll +O
Apply KVL on loop fcbef12- 0.075t, 2tL =0l, =i60-26.671 +A
I
Apply KCL at node'b':
I +1, =lL +O
Subslitut€ Eq.1 and Eq.2 in Eq.3:240-4011 +160 26.67|1 =lL67.6711 = 400
lr =5.91A
'=#=oApply KVL on loop'odbacr ./J0r,
u-,lE) 'lt)=
ol3I l2 ,/
v 0.167rR=0-o
Substilute Eq-1 To Eq.2:
v_o j 67fl!q)(R) = otRr',v - 25.05 = 0
V =
25 volts
q = 0.3(4) = 1.2 O
f, =0.3(6) = 1.8o
UseNodalnode method
ApplyKcLatnode'a: 6oov
lr +1, =lL600 - vr 560 - vL -^^
1.2 1.8
Lr s. ;(0.2) - rr (3) = 0
r5 3.02tr =0rr . 0.4086
rr :0 5A
,TI
t =4.2O
+
J]L
IIL
I
+
.F
Solutions Test 3 7776 l00l Solved Prcblens in Electticol
Use Thevenin's lheorem:Solve lor looking back resistance across the load lerminals:
Note Open.,rcL i P . )Lror-.0, d i'lae lwo I dre.ies. B/ r..p+ro.'F. l
ba.k rerstan.e F equalto the equ valent rcti(ance oftwo rdentical re5ittois
r 15"22
Solve the open circuil voltago;crcss the oulput.
Note, Since the batteies are identi.al, tltre j5 Do circulating cuft.nt i'r thefor the two batteres, thu, the open circuit voltage is equal to ile open civoltage oF.a.h battery.
Ea =24v
Using th6Thevenin s equivalent circuit:
'.-**;=;#"=o,,,PL = IL'RL +O
Substitule Eq.1 to Eq.2:
',*={ 'o I'n,\0.75+RL, '
ro 75*R, )t = 57ffit
100
0.5625+1.5R1 +Rr' = 5.76Rr
R.'?-4.26R. + 0.s625 = o
U6ing the quadratc fomula:
- +.20.,,{+:of -otrxosozsl' 211)
RL(.) =412oRrl-) =0.136o
4.26 r3.9A72
U.. Nod.lnod. mclhod:Itdy KcL .l nodo 6lr'|ts-lrlrn vr ,122 V1 _ Vr
0, 0.3 0.5
It llv, ' 10 66 3.33V1 = 2Vl
l0 vl - 103 88
Vr . l0 03 volle
. t2 6 V, 126-10.03tt'oz'Tlr '12 E6A
Uta lhavcnin'3 lheorem.0t n ckcull RL and solve for Ro
2+Ris-R,, -!-Or,- l.;oaUldllul. R"in Eq.li
sr-2'213*ut-zo
Udng Thov6nan' s equation :
i-n"rh,
lli, -2olr-t' 1no *n.1= 312121
lu - tZ votts
a l) r€sislor is insert€d
-2r4=6OEo 12- R"tq- 2+6
-15A
1
aIiIr
ll
a=l-54 . rt:l 59
Solutions ?est 3 797t l00l Solved Ptoblens in Elecoical
t,=005xr, = 0.05(4 x)
Apply KCL on node 'b':lr +1, = 300
11 = 300 - l, +(D
Let VL = voltage across th; tain
Apply KVL on loop dabed':420-11(0.0sx)-VL =0vL = a20- l(0.05x) +O
Apply KVL on loop fcbef:alo - lr(0.05)(a - x) vL =0V! = 4t0- lr(0.2- 0.o5x) =)O
SubslMe Eq.1 in Eq.2:V! = 420 - (3cro - 1, X0.05x)Vr = 420 - 15x + 0.05x1, +O
Lqualo Eq.3 to Eq.4:11O-0.22 +O.05x12 = 420 - t5x + 0.05x1,
, 410 -420 + 15x,,- 02
l, =75x-50
Subsiilute 12 in Eq.4:VL = 420-15x + 0.05x(75x -50)Vr =420-15x+3.75x'z -2.5xvt = 420 17.5x+ 3.75x2
!L=6-175*75'0=7.5x-17.5x = 2.333
l, = 75x-50 = 75(2.333)-50lz=125p't1 = 300,t' = 300 125
11 =1754
h.0.0.06(6)-03o
tlo loorr -E, -loor, =o
tlo..loq1.35)-E, -1oo(0 3) = o
l' . 6tl v
n'l'fl'*=#"*i. tlttl||. llodrl nod. method:
Arrly KcL d nodo '4.
l. rlr - llu. v 8-V. ,
2la-0.0v1 + 8-vr =ll| .11- l.5v! +lD
t.ft-1F=olqud. Eq.1 lo Eq 2:
7 656la 1.5v1 =-
llvr 1.5v.' = 7.656
v,'s::v,*s.to+=o
Ulng lhe quad.alic fomula
I 33,Jtg.33)) -4(1)(5.104) 9.33:8.162u'' -' o;- = z
V.(')=875volts
V| = o 584voltg
i-027(6)-1.36n
' t!!-
t0 l00l Solv.d Ptoble^ in F.le.bical $lutit' Te 3 EI
fl"l, l!t ,,l.lrr.t!n) Pow.r transfcr' lor,l rcs rtancc mu5t be cqu vrLcnt t" th'l,i,l ,!tl.r, !,.nd.rD.. meilre.l ri the Lo!,]terDinalt
Oll.n clrcuil Ry and solve lh€ looking back resislance across its terminels
2(8) ".^2'8-Ro - 1.5 O
r20v
N'(r 1,, n)itimom Power van56r, load rc5istance must be equN€lent to th.h'lrnt h.rc[ rotstance measu'cd at the load teiminaG
Substitute VL = 8.75 in Eq.2
1, =Z!!9=e6754" 8.75
n=!'-='875rL 0.875
p r,-13r=e.zse
n,,
llr
Apply KVL on loop abcdefa':12.5-tb?)-12 = 0ls = 0.25A
Apply KCL at node'b:lc = lr +lb = 6.25+0 25
L = 6.5 A
,-29
R:20
I2I/
i
Note, Fofmarrmum powei tranrFer, loa,l resistance must be Eoi€lenttoth€looLing back rctrtance mearurcd atthe loadteminals. ln thrs problem, tu = r
r,-., -z[ffi'zsoot)-039(l
E.Rh* +Rr 0.39+ 0.39
- 760.23 A
- t, rR,
= (769.23)'(0.39)
.230 77 kW
E"= 600 vRL =Ro =r=00'lo
Using Thev6fl in's equation:
r. =uft.=ffi"=ooonPL =r!'?Rr = (600)r(0.01)
Pr = 3600walts
RL =Ro = r +Rlr =05+9.5=10O
Using Thevenin's equalion:. E. 12Ot =
Rn +R, = 10+10 =oA
Pr =rr?Rr = (6),(10)
Pr = 380 watts
IIL
R1
IrI,
taft
600
Note, For maximum pola$tianjF.r, Joad rcsittance D{]5t be equivalentiothclooking back rcristarce mcasurcd at the load tem in.ls tn th rs problem. Rr = r
Nrrt. l(r'naximump ertransFer,load rcsittance must be equie lcnt to thc
Lr,Liir,I'i.k rcsistance measured atthe loa,l teminals
Otan ckcuit Ry and solve lhe looking back rssistance acloss its terminalg.
12v
ir
&
t:0.011)
R,,-=9.50
r=0.5n
120
-o- = !?1]!!l = o zoa o' 0-25+1.25
RL=fu
E2 1001 Solyed Prcbtens ih Etectricat
n.=ozs'fffffi-orzeor,=-;=oi;.28.044
By CDT:
, l, (1.25) 28.O4t1 25\'L
= i25E = -E i2os = 21 04 A
P, - t.zRL - (24.0a),(0.208)Pt =12O21W
Noie,.ln,l€pendcnt.of.be cn-u rt .on n.d roD. rhe toial powe..l,awD by aequal lo the sum o{ th. powe6 td!f,D b eaLh rcr;stqnce.
Lel: R = resistance of eadt leai3ld
R = Pr +P2 +P3 +........+p1,
,,R, =6(*)'R.6(*)'R
rn, =f(,n).${,n)
R. =:R +:R93AR. =9R=!(a86Rr = 1.66 o
.t a,ra -['-z(1)1"=f n-6lor tha vollagc 6cros6 branch 'ab:
lor th. tot.l curr€nt:
.rR, '.Ogq 4 to Eq.3:
rl)rl
:q 1 ln Eq.2
\ -0t, -0
Itfi
!
r1
12
t-
Il^
I
I3
!6
l,.t,Ri.(4Ox10'Xs)lr - o.2o v
/,1
+60wr2v
t
tt2
.1 ,2,12'
.! 167 Q
lda {. thc P oblen #99 ro, 'l e'o'-)ur. 'o <o've /or P
I R . roalltence ot each resistor
.6a - 9ror=sn68
Solutions Test 3 83
@ Lel: R = resistance of each rosistor
Apply KVL on toop ,abcda':
(r- ztj)R - tlR - (1 _ t')R - trR = 0lR 2llR - 2lrR -trR + t,R =0 ,r-Srr+ru =o -1,l, = 511 +l=O
Apply KVL on toop ,efghe':
(, l,)R irR 2trR t,R=olrR r,R t,R 2t,R t,R = 0 I._,st2 \ = 0-@
O,K IIJ d
*-t,I-2r lIt 'tn1".
I
\
itat
) "'+r''
y'
_lj,.,.,l
,l L4at21V
E4 I 00 I SobeA Problens in Elect.itul Snlutions Test 3 Es
SoMng for the resistanc€ of the lamp:
R, =:.L=\:L=21c'Pl 80
UBo Nodal node malhod:Apply KCL on node'e':ll =1, +13 +l.i
24,V V -12 V V1 0.1 4 21.
24- V = 10V -120 + 0.25V +0.4'l6V11.666V = 144
V ='l2.34volts
Use Nodalnode melhod:Apply KCL at node'a':
130 - VL 120-VL Vi3240
a3.333-0.333V! +60 0.5Vr = 0.025V!0.858Vr = 103.333
VL = 120.43 volts
| 120 12O 43'r- 2-lb = -0.215 A
Notc: Tte negativ. sign denotes only that the actual curcnt is oppo5jtegiven dircctron (charging the batt€ry).
Apply KVL on loop ebcfe:120+r;Rr-V=012o +211)-v =0V =122volts
Ndt., 5iDcc thc voltmet€r has a very L'igh rc5itta nce .omP?red to the resistanc's in
it'. "r'*'t.
tt'" -"*t no* tow.rds th; voltmeter is negligible
12'' 20+R
'---lL=o2A'| 1o+50t2y
Apgly KVL on looP 'cdabc':
O 435 + l,R -lr(10) = 0
ocu.(jfo)n-to,x'or=o
-1?L = t.ses20+R12R=31.3+1.565R10.435R = 31 3
R-3()
L,so Thevenin's theorem
Not. Shoit circuit the 12-V suPPly' oPcn crrcuit the 5o-fl resrstor and solve tle,)(.l{ln9 baak resirtancc ofthe.rr.uit at thesat€minah
r,.f -$-...'oArply KCL rt nod.'b'll,.l' +h - 2+ a0.67
tr - a2.67 A
Atply KVL on loop dabed':
a. - 12.g7ll) -2(1, - 12O = 0
lr .164 87 volts
,t+
f"30
Es
II30v
Ich
+
v
I
{Ir-24+- 120v
ta lo
Solt uuns Te 3 8786 I00l Sol'ed Ptuhlens ib Electti.al
R^ = !!{M* lq9qr" 20+40 10+30Ro = 20.833 O
Solve forthe open circuit voltage:
r.= 12 =o.ze
t" = 12 =ogl' 10+30
Apply KVL on loop'dcbad':Eo +r,00)- tr(20) = 0
Eo = 0.2(20) 0.3(0)Eo = l volt
Ljsing the Thevenin s equivalent circuit:
Ro +Rr 20.833+50
l=0.014A
FlmIllly lh€ parall€l
,. - '16\ -zqa'' 4rB
', - fa) '' 0"
f rrllrlorm lh6 currenl sources lo their equivalent vollage sources:
t,-0(2.4)=14.4vl, - 10(1.6)= 16V
!0 ll1 6+1+24)-1a4=0ln 5 0
l'032A
l2v
Transform the voltage sources lolheir equivalenl current sources.
R..draw the circuil:
Apply KVL on loop dabcd'i(l^ ls)2 lB(8)- (lB + lc)2 = 0
N,n. lA = 6/4r lc = 34
Sub6litute:(8 ls )2- 8lB (B + 3)2 = 0
12 2ts 8ls-2lB-6=olr.0.5A
v rB(8) = 0,5(8)
200 10n
-+10n 300
It I 00 I Solyed Prcblens in Etectricat
solve th6 looking back .esistanceOpen-circuit lhe 1-ohm r6sistor andcirclil al these terminals.
Solutions Test 3 89
9p.fi olroull lhc lo-ohm rosistance and solvo the op€n circuit vollage atthese
Irn nal.
Note, T1'e cu[ent source i5 open€d wbilethe voltage soorces aie short_ci
By insp€ction, Ro = 4C)
Shorl-circuil th€ 1-ohm resistor and solve the shorl circuil cur€nt thatto lhis branch.
t)a "lNorton k equival ent c itcu it
Apply KCL at node 'a':
^_t -2O-12
t , the Nonon's equivatent circuit:
, t*Ro l2)t4,''-R"*&=4;l! =164
50
o n, (l) toa
Afrt KVL uBino the loop as shown :
v (0lv,x5)-10 = 0
vr 0 6vr 10=0
Vr - 20 vottr
lr-Vr _20volls
ahorl clrcult lho lo-ohm resistor and solve the short circuit cunenl
l1_
h=0
Th ev enin's equ iY ol e ht c it cu t
Nol. lh. volt.ge-controled.urrcnt source a' sho,n will delivei zeto cu*ent
Alply KvL using lhe loop as shown:
l0 l.(o+ 5) = o
lr'tlr- - E-q=?9=zoo
Ua. lho Thevenin's equivalenl circuit:
r. - El=-?L=6667n'| R" +R' 20+10
vt
t0r
IIL
ir'
90 1001 Solyed Prcblens tn Elec|ical
Notc: When a so!r.€ D opefating alone, the other c! reDt Fdrces mun bcwhile the oth€r voltage sou rces must be shod-circuited.
Lel, the vollage source lo opersle alone:
2-ohm resistsnco and solve lhe open circuit voltage across
------->I -----> I = 0
Sol rio6 Test 3 9I
0llrlr rtrcull lho
+
+
6t)tatv,
6r/
6+v1-t(6)+3V=06+4V, =61',+O
V =I(1)=r+A
Substitule Eq.2 in Eq.1:6+4(l')=d'
l=9=se2
Lel, the currenl souce to operal€ alone:
Apply KCL at node'a':3 = 11+ f'
3 = 1.667V
1=1!-=1[9)=12a66
l= l'+ l"= 3 + 1.2
t=42A
Notei since the drection of cureots for botb I' an,lrcsultant cu"ent that will 6ow i,r the 6-ohm r6irtance,5!r, ofthetwo cunentyeipectively.
N.il. ! r!c tciminah 'ab' E oPcD-cir.uited (l = o), the current controlled curent
|.rn,. wrl,lclver sro cuifent (equi!4lenttoan oPen circuit)
lf ln.p6clion, Eo = 30 V
llrorl clrcuil the 2-ohm resistance and solve lhe shorl circuit current
ApDly KcL at node 'x:l,-lr0.6ll| - 1 .8r
to v, ., arv. J- 6 - -l 4Jt 02v, =0.4vrV, - l0 volts
r, -r !:=f =zsr
..-i:=#1,, - 12 {}
U16 Nodalnode method:
"""'> 1' x -"'t'l = I.
92 1001 Solrc.l Ptablens ib Ele.biul R.R
Apply KCL at node 'a :
6 = l+ li
6= % r k3:13246 = 0.5V. +0.25% + 0.51- 3
n=orru *ool]'l" \2 )I = 0.75V, +0.25V.
t=:a=:=45A22
lJse Nodal node methodl
I2V
Apply KCL al node 'a'l+4=li+414=11+31
r=Y*"ftz-v'1
4=0.25V+9-0.75V0.5v = 5
V = 10 volls
G0u10 Bsuw
tlrrl l.yv ol €l.cttGt tic6: Like chaQes repoleach othet and unlike
a[r!.. rllract sach othet.
aaoonrl l.w ot olocuostatlca: The lorce of altracion or repulslon between
ir]u.r 'i.jrocily proporlionalto lhe prcducl of lwo charges and inversely
Fq';'t 'trlto lh€ square ofthe distance b€tween lhem
rJ, (lr r charge in each tody (coulomb)
l " -.I,lolule P€rmitlivity- LE54 x '10 t'zfarad psr meter
l, - rolalrvo pemittivity or dielectric constant- t(ono), for free space
rl - (lritdnc€ between the lwo bodies (meleol - ",slantn Sl unils equalloI r10r
b'',.rr,,', ".1)"e. O =.hr -ou omb. d ".entmete .n,lri.r,,,l {, I, n a I coulombofcharqe
tr[CIf,OSTING NIttI Ilo potlntlal-the electdc potential resulting from the location ofcharged
V - .ln(lrostalic potential (volt) at a distance ofdl!rly ol charue Q (coulomb)
(meter) from a charce
