PRINCIPLES AND PRACTICE OF ENGINEERING ELECTRICAL AND COMPUTER ENGINEERING BREADTH MORNING SAMPLE TEST SOLUTIONS

Transcription

1 PRINCIPES AND PRACTICE OF ENGINEERING EECTRICA AND COMPUTER ENGINEERING BREADTH MORNING SAMPE TEST SOUTIONS Slutin (and Tet) by Kenneth. Kaier, P.E., Ph.D. Prfer ECE Department Kettering Univerity Flint, MI Diclaimer Reanable effrt have been made t publih reliable lutin fr the ample tet, but the authr cannt aume repnibility fr the validity f all lutin and fr the cnequence f their ue. K. Kaier 006 Verin 04/13/07

2 101. Anwer i (C) The tandard gain equatin fr an inverting ideal amplifier i R f Av Ri where R f i the feedback reitance and R i i the input reitance a hwn in the fllwing figure. R f V i R i V frm fr Bde pltting, fllw (auming ( ) ( ) 1 1 Thi exprein can al be ued fr inuidal teady-tate cnditin when the feedback and input cntain, C, and R 1 R jω C 1 1 R R R R jω C jω C 1 jω RC 1 jω RC jω C1 Av jω C R 1 1 R1 R1 R1 jω C1 jω C1 jω C1 jω C1 jω C1 R 1 jω RC jω C1R jω C1R 1 jω C1R1 ( 1 jω RC ) ( 1 jω C1R1 ) ω ω 1 j 1 j 1 1 RC R1C 1 A quick partial Bde magnitude plt f thi exprein, which wa written in tandard 1 R C < 1 R C. Thi i clearly a bandpa filter with tw cutff r break frequencie. The anwer i (C). 0 db/dec. 0 db/dec. ω 1 1 R C R C 1 1 K. Kaier 006 Verin 04/13/07

3 I there a fater methd f determining thi lutin? Ye. Recall that at lw frequencie (F) an ideal capacitr appear like an pen (r nearly ), and at high frequencie (HF) an ideal capacitr appear like a hrt (r nearly ). Thu, R f R R R f R 0 0 AvF 0, AvHF 0 Ri R1 Ri R1 0 R1 the gain at bth lw and high frequencie i zer. The nly reanable cncluin i that at me frequency between lw and high frequencie there i me gain. Hence, thi mut be a band-pa filter (BPF). Thi quick methd can be applied t many ther pamp circuit perating in inuidal teady tate. 10. Anwer i (B) When wrking prblem f thi type, I alway redraw the circuit that the aertin level between gate and input and utput are matched. Thi help eliminate the need t apply DeMrgan law (directly). I begin with the utput and wrk tward the input(). Since the utput i high aerting, I firt draw an equivalent gate fr the utputmt gate, which i al high aerting: A C B A E Cntinuing tward the input, equivalent gate are ubtituted int the circuit when required that bubble meet bubble and nnbubble meet nnbubble: A ( H ) C ( H ) B ( H ) A( H ) E ( H ) The high aertin level are indicated in thi lat figure. In rder t match the lwer C H C : bubble input f the upper OR gate, rewrite ( ) ( ) A ( H ) C ( ) ( ) ( ) B H A H E ( H ) Nw that all input and utput (including the f the gate) are matched, the lgic exprein fr the utput i written dwn by inpectin, which i equal t (B): E ( A C ) ( B A) A B C K. Kaier 006 Verin 04/13/07 3

4 103. Anwer i (B) Stability prblem f thi type are eay t lve a lng a the Ruth Tet i undertd. Fr negative feedback a hwn, the cled-lp tranfer functin, the rati f the utput t input, i K ( 3) C ( ) G ( ) G ( ) 5 lp R ( ) 1 G ( ) H ( ) K ( 3) Where G() i the tranfer functin f the frward path and H() i the tranfer functin f the feedback path. (Carefully Nte: If the feedback wa pitive, then the denminatr f the cled-lp tranfer functin wuld be 1 G ( ) H ( ).) The Ruth Tet require the ue f the characteritic equatin f the cled-lp tranfer functin. Thi i btained directly frm the denminatr f the tranfer functin after me implificatin: K ( 3) K ( 3) Glp ( ) 3 5 K 3 5 K 3K ( ) ( ) 3 The characteritic equatin i 5 K 3K. Frm the cefficient f thi characteritic equatin, the Ruth Tet can be applied: 1K53K 5 K 3 K 03K 5 The ign f the firt clumn element huld be the ame fr tability. Therefre, 5K 3K > 0 and 3K > 0, and the anwer i (B) Anwer i (C) (A) i nt likely t be crrect ince capacitance i added acr inductive lad (uch a large HP mtr) t reduce the net reactance f the plant. Thi i ften dne t reduce the current n the pwer line leading t the plant. The pwer cmpany i cncerned abut thi current ince it i l f real pwer, due t a I R le, that they are nt meauring with a real pwer meter at the plant. (B) i nt likely crrect ince capacitance i ften added t reduce the verall reactance and hence reduce the reactive pwer in VARS. (D) i nt likely crrect ince the paraitic capacitance f a mtr i nt uually cnidered a urce f pwer l (ideal capacitr d nt abrb pwer). The anwer i bviuly (C). The pwer factr (PF) i defined a cθ where θ i the phae angle between the vltage acr and current int the plant (r any lad). By reducing the net reactance f the lad, by canceling inductive reactance with capacitive reactance, the lad lk mre reitive and thi phae angle i cler t zer degree. Thu, the pwer factr i brught cler t ne and, in the language f electrical engineer, the PF i imprved Anwer i (A) When teaching electrnic, I like t ak quetin invlving ideal p amp and uperpitin. In thi circuit, imagine turning ff v (i.e., replacing thi vltage urce with a hrt circuit). In thi cae, the circuit reduce t an inverting amplifier (R 3 and R 4 K. Kaier 006 Verin 04/13/07 4

5 huld nt affect the circuit much ince zer current pae int the input f the ideal p amp) with a gain f R R Av 1 v 1 v1 R1 R1 Next, turning ff v 1, the circuit reduce t a nninverting amplifier. Fr a tandard nninverting amplifier the gain exprein i R Av 1 R1 R4 Hwever, the vltage int the input f the amplifier i nt v but v v, which R3 R4 i btained uing vltage diviin. Therefre, the utput vltage due t v i R R4 v 1 v R1 R3 R4 Uing uperpitin, the ttal utput vltage i R R R4 v v 1 v v1 1 v R1 R1 R3 R Anwer i (A) GFCI (r jut GFI ) are imprtant device that are ften ued in utlet near water and in garage, baement, and utdr. (See the electric cde fr detail requirement.) Briefly, GFCI wrk by detecting an imbalance in current between the ht and neutral cnductr. If thee current are nt the ame, then current mut be returning back t the pwer urce via me ther rute (e.g., the uer f the prduct cnnected t the utlet). GFCI, hwever, d nt prtect an individual in all pible electrical-hazard cenari (ee my fabulu reference bk Electrmagnetic Cmpatibility Handbk fr mre infrmatin n thi and many related cncept). Circuit breaker, a tated in anwer (C), are typically ued t prtect equipment and help prevent cnductr frm verheating. GFCI (fr typical reidential utlet) trip arund ma, which i much le than the typical tripping current f 10 A r mre fr circuit breaker. The let-g current i between 6-9 ma at 60 Hz and i a functin f many factr. Thu, (D) i al crrect. Circuit breaker are placed in erie with the ht cnductr and pen r break when the current in thi cnductr exceed me level. Circuit breaker d nt ene an imbalance in current between the ht and neutral cnductr a GFCI d. Hence, (B) i al crrect. (A) i nt likely crrect ince circuit breaker ften pen r trip lwly nt t quickly Anwer i (B) Ideal reitr abrb real pwer and a wattmeter meaure real pwer in watt (W). v t, then uing vltage diviin fr the tw reitr in If the ignal vltage i given a ( ) erie, the vltage acr the Ω reitr i v t 5 v t ( ) ( ) K. Kaier 006 Verin 04/13/07 5

6 If the urce vltage wa dc, then the pwer abrbed by thi reitr wuld be equal t V dc 5 P If the urce vltage wa inuidal (r ac) with zer dc ffet and an amplitude f A, the time-average pwer abrbed by thi reitr wuld be 1 A Vrm 5 5 P A where Vrm fr ine wave with zer dc ffet where V rm i the rm value f the inuidal wavefrm. The rm value fr ther peridic wavefrm with a perid f T can be btained frm the general exprein 1 T vrm v ( t ) dt T 0 Unle the wavefrm i relatively imple, it i prbably reanable t aume that n multiple-chice tet f thi type, inufficient time i available t apply thi exprein t btain the rm value f a peridic wavefrm. The rm value f the triangular wavefrm f even-ymmetry and dc ffet given in thi prblem i given in my (and many ther) handbk a 1 3. Therefre, the anwer t thi prblem i (B): 1 V rm P W There i al anther way f lving thi prblem that de nt require the ue f the exact exprein fr the rm value f thi triangular wavefrm. By inpectin, the average value f the wavefrm i 6 V (add up the ttal area and divide by the perid). Therefre, a fat etimate fr the pwer i 6 5 P 1.5 W The clet anwer i again (B). The rm value f a wavefrm i alway greater than r equal t it average value: 1 1 T T vrm v ( t ) dt vavg v ( t ) dt T T 0 0 Becaue the anwer are well paced, numerically, thi methd wrk. When taking multiple-chice tet, alway lk at all f the anwer (and their ign if apprpriate) Anwer i (D) The riginal ct f $55K and the future alvage value f $0K huld bth be cnverted t an annual value at a 6% interet rate: K. Kaier 006 Verin 04/13/07 6

7 ( ) ( ) A 55K A P, 6%,5 0K A F, 6%,5 55K K K An interet table wa ued. The clet anwer i (D). Nte that the anwer i nt 55K 0K 5 7K. A gd multiple-chice tet will have many f the wrng-way ( ) lutin a chice, alway carefully think thrugh the prblem and duble check Anwer i (C) Thi prblem include bth a mnthly and an annual ct. The mnthly interet rate can be btained frm the annual interet rate by uing the equatin ( φ ) i 1 1 where i i the annual interet rate (a a fractin), φ i the crrepnding mnthly interet rate a a fractin, and k i the number f mnth in a year. Thu, ( ) ( ) φ φ φ φ The mnthly interet rate i 0.49%, which i cle t 6%/1 0.5%. In me prblem, thi mall difference may be imprtant. The preent value f the mill can nw be btained by adding the mill initial ct t the preent value f the mnthly and annual maintenance ct: P 0, P A, 0.49%, 7 1 1,500 P A, 6%, 7 k 1 1 ( ) ( ) ( P A ) ( P A ) 0, , 0.49%, 7 1 1,500, 6%, 7 Nte the number f cmpunding perid fr the mnthly rate i nt 1 but 84. My table did nt have 0.49% interet rate r 84 cmpunding perid. S the actual frmula fr P/A will be ued: ( P A, i, n) n ( i ) i ( 1 i ) ( P A ) 1 1 n 84 ( ) 1 ( ), , If the 0.5% table i ued with n 85, the reult i 69.11, which i cle t that btained. The anwer i (C): P 0, , $49, Anwer i (B) The ynchrnu peed f an inductin mtr with p ple perating at a frequency f f i 10 f n 1,500 rpm p 4 The lip, a a fractin, i defined a n n n where n i the actual peed f the mtr. Slving fr n, n n n n 1 1, , 460 rpm ( ) ( ) K. Kaier 006 Verin 04/13/07 7

8 The full-lad rtr peed i le than the ynchrnu peed. Althugh the rtr field rtate at the ynchrnu peed, the rtr itelf de nt Anwer i (D) Thi i a tandard K-map prblem that huld have been cvered in all intrductry cure in digital ytem. The K-map i ued t help rganize the relatinhip between the input and the utput and find a minimal exprein. The K-map fr the utput E btained directly frm the given truth table i AB\C A um-f-prduct (SOP) utput exprein i btained by gruping tgether the 1 : E AB BC A prduct-f-um (POS) utput exprein i btained by gruping tgether the 0 : E B ( A C ) which, when expanded, i the ame a the SOP exprein. By inpectin, the anwer i (D). Thi can be eaily cnfirmed by analyzing circuit (D): A ( H ) A ( ) A ( H ) B B C AB BC E AB BC 11. Anwer i (A) The tw extreme utput cnditin exit fr a BJT when the tranitr i in cutff and aturatin. When cutff, little current pae thrugh the tranitr frm the cllectr t emitter, and the tranitr witch i pen. Bth the EB and CB junctin are revered biaed. Thu, a high impedance i een lking int the emitter and the vltage acr the lad reitr i btained uing vltage diviin: v 15.5 V min 10 (A a imple exercie, check t verify that the tranitr i actually in cutff fr the given et f reitr.) When aturated, the tranitr witch frm cllectr t bae i fully cled and the vltage frm the cllectr t emitter i mall at value f 0. V. The EB K. Kaier 006 Verin 04/13/07 8

9 junctin and CB junctin are frward biaed. Thu, the vltage at the emitter i 0. V le than the vltage at the cllectr: v max V The anwer i (A) Anwer i (B) Thi i a erie reliability prblem. The prbability that the tranmitter will nt malfunctin (nt fail) i 97%: P T ( ) The prbability that the receiver will nt malfunctin (nt fail) i 95%: P R ( ) Therefre, the prbability that bth the tranmitter and receiver will nt malfunctin i (B): P P T R Thi i equal t 1 ( ) and i le than the reliability f either the receiver r tranmitter alne. Fr a parallel reliability prblem, the reliability huld be greater than either the receiver r tranmitter alne. In thi cae, if it i acceptable t have either r bth the tranmitter r receiver perating prperly, the reliability i 99.9%: 1 1 P 1 P ( ) ( ) T R 114. Anwer i (D) In thi prblem, the tw event that can caue the amplifier t malfunctin are nt mutually excluive. That i, they are nt independent f each ther. Thi can be een by umming the tw prbability f failure: P P Famp Fcap Thi i nt equal t the verall prbability f failure given a Since the tw event (i.e., prbabilitie f failure) are nt mutually excluive, there i me verlap in the Venn diagram fr thee event. The verall prbability f failure i then given a Pttal PFamp PFcap PFamp& cap PFamp& cap P Famp& cap There i 0.05% chance that the audi amplifier will fail due t a malfunctin in bth the final pwer tranitr and the high-vltage capacitr. If the event were independent f each ther, P & 0. Famp cap 115. Anwer i (A) The intenity f the electric field between tw cnductr at different ptential generally increae a the applied vltage acr them increae and a the ditance between the cnductr decreae. In thi pecific electrde cnfiguratin, the intenity i greatet directly belw the cylindrical cnductr where the harpne f the cnductr i the greatet (r where the radiu f curvature i the mallet) and where the tw cnductr are the clet. The directin f the electric field at a pint f interet between the cnductr i determined by placing a mall pitive pint charge at thi pint and K. Kaier 006 Verin 04/13/07 9

10 qualitatively determining the directin f the frce n it. (The pitive charge i mall that it ha little effect n the applied field frm the charged cnductr.) When a negative vltage i applied t the circular cnductr relative t the grund plane, the grund plane i pitively charged and the circular cnductr i negatively charged. The mall pitive tet charge will be attracted t the negatively charged circular cnductr and repelled frm the pitively charged grund plane. (Recall that like charge repel and unlike charge attract.) Hence, the field at lcatin A i in the directin f the circular cnductr Anwer i (D) Thi i a favrite prblem f mine that I have given (in variu verin) t my tudent. Firt, it i imprtant t remember that ideal rdinary tranfrmer d nt pa dc current. (Dc current, hwever, can affect the magnetic cre f a real tranfrmer.) Thu, the dc vltage n the primary r input ide will nt pa t the utput. Secnd, nte that the primary r input ide f the tranfrmer are cnnected in parallel the ame 340 V peak-t-peak vltage i applied acr bth primarie. Third, ntice that the ecndary r utput winding are cnnected in erie adding (a determined by the plarity marker). Thu, the utput vltage i V peak-t-peak The rm value f the utput i therefre (D): V rm 10 In the figure that fllw, the dc ffet acr the primary ide i nt hwn and the rm value f the wavefrm are hwn. 10 :1 10 V 1 V 10 :1 4 V 10 V 1 V T help build yur cnfidence with plarity marker, tw ther ituatin are hwn with the ame input vltage. Fr the 0 V utput vltage ituatin, the ecndarie are cnnected a erie pping r bucking. K. Kaier 006 Verin 04/13/07 10

11 10 :1 10 :1 10 V 1 V 10 V 1 V 10 :1 0 V 10 :1 4 V 10 V 1 V 10 V 1 V 117. Anwer i (D) An auttranfrmer can be cntructed frm an rdinary tranfrmer by cnnecting ne ide f the primary t ne ide f the ecndary. Hence, dc ilatin i lt: (C) i crrect while (D) i nt crrect. DC ilatin implie that DC current (r vltage) cannt pa acr the device. Baed n hw the winding are cnnected, the utput vltage can either tep up r tep dwn the input vltage a hwn in the given figure. Anwer (B) i therefre crrect. Finally, with auttranfrmer a winding i hared between the input and utput, and in the given figure thi crrepnd t the cil with N 1 turn. _ N N v v v v v 1 v v v 1 v 1 N1 v 1 N1 tep-up auttranfrmer additive plarity tep-dwn auttranfrmer ubtractive plarity 118. Anwer i (A) When qualitatively analyzing paive filter, fur prpertie huld be recalled: 1) an ideal capacitr ha a reactance f X c 1 ω C ; thu, a capacitr appear like an pen at lw frequencie and a hrt at high frequencie ) an ideal inductr ha a reactance f X ω ; thu, an inductr appear like a hrt at lw frequencie and an pen at high frequencie K. Kaier 006 Verin 04/13/07 11

12 3) a erie C circuit appear like a hrt at me renant frequency 4) a parallel C circuit appear like an pen at me renant frequency Uing the firt tw prpertie, it i clear that at lw frequencie the utput vltage i abut V R A R R A high frequencie, n the ther hand, the capacitr hunt and the inductr blck the input ignal. Thi i a lw-pa filter. Althugh meh r ndal analyi can be ued t determine the exprein fr the utput vltage (by lving three equatin with three unknwn), a ne-line lutin can be quickly written dwn by uing vltage diviin twice: V V R jω R j C ω jω C1 jω C R R j jω R ω jω C jω C firt vltage divider baed n element a and b jω C ecnd vltage divider baed n element c and d element a element c V R C1 C R V V R C1 C R V element b element d After much careful implificatin, the previu equatin can be reduced t anwer (A). Obviuly, I deigned thi prblem t be wrked in a few minute (nt 0 minute r mre). What fllw i the trick lutin that can ave yu a lt f time. Of the fur anwer prvided, nly (A) and (D) have the crrect lw-frequency gain given previuly (let ω 0 in thee equatin). (A) i prbably mre likely than (D) ince thi i a thirdrder filter (i.e., cntain three eparate energy trage element), and the denminatr in (A) cntain a ω t the 3 rd pwer Anwer i (D) The exprein z(t) i repreenting the tandard amplitude mdulated (AM) ignal with a carrier: z ( t ) A 1 km ( t ) c [ π fct ] Ac [ π fct ] Akm ( t ) c [ π fct ] carrier ideband Hence, x(t) i an AM ignal withut a carrier al referred t a duble ideband (DSB) ignal (r AM ignal with a uppreed carrier). Therefre, the anwer i either (A) r (D). There i nw a 50% chance f gueing the right lutin. The exprein y(t) i K. Kaier 006 Verin 04/13/07 1

13 repreenting a phae mdulated (PM) ignal. The phae f the ignal, ften een in textbk a θ, i changing directly with m(t): y ( t ) Ac π fct km ( t ) { θ The anwer i therefre (D). Unle yu have me backgrund in cmmunicatin thery, it may nt be apparent why (t) i repreenting a frequency mdulated (FM) ignal. If the mdulating ignal wa ummed directly with the carrier frequency then it may be mre bviu why the frequency i changing directly with m(t): { π c ( ) } Ac f km t t. Hwever, thi i an uncmmn way f repreenting an FM ignal. It i mre cmmn t talk abut the intantaneu frequency f the ignal defined a the time derivative f the argument f the cine: t d π fct k m ( λ ) d λ π fc km ( t ) dt 0 Thi intantaneu frequency i directly prprtinal t m(t) and i thu FM. 10. Anwer i (C) A cmmn way f repreenting an FM ignal i t x ( t ) Ac π fct k f m ( λ ) d λ 0 where k f i the frequency deviatin cntant in rad/ec. Fr a inuidal mdulating m t a c ω t, the mdulatin index i defined a ignal, ( ) ( ) m m k f am β ω m Carn rule ue thi mdulatin index t determine an (ften fairly gd) etimate f the bandwidth (in rad/ec) f an FM ignal: k f am BW ( β 1) ω m 1 ω m ( k f am ω m ) ω m A een frm thi exprein, the bandwidth f an FM ignal i a functin f the frequency deviatin cntant, amplitude f the mdulating ignal, and mdulatin frequency. Subtituting the given infrmatin int thi exprein, the deviatin cntant i equal t 3 3 ( ) π ( ) π π ( 5 10 ) k f am π ( ) k f a a (β, the phae cntant, i 0.5 in thi cae, which implie the FM i narrw band.) Uing a prime t repreent the new bandwidth, amplitude, and frequency, m m K. Kaier 006 Verin 04/13/07 13

14 3 3 ( ) π ( ) π BW ( k f a m ω m ) am π 5 10 am am The anwer i (C) { π ( ) π ( ) π ( )} π ( ) 3 ( ) 11. Anwer i (B) Thi i a firt-rder circuit with a ingle time cntant, τ, equal t R eq C where R eq i the reitance een by the capacitr after the witch i cled. When determining thi reitance, all independent upplie are turned ff: independent vltage upplie are replaced with hrt and independent current upplie are replaced with pen. The equivalent reitance i given by inpectin a ( R R1 ) R Req ( R R1 ) R R R1 R Baed n the time cntant prvided, the nly pible anwer are (A) and (B). The general lutin fr any current r vltage in a ingle time cntant circuit i A Be τ Even if me ther quantity i requeted, I generally will lve fr the vltage acr the capacitr (r current thrugh the inductr fr an R circuit) ince thi vltage cannt change intantaneuly. Thi capacitr vltage i then ued, if neceary, t determine ther quantitie uch a the current int the capacitr. Initially, the vltage acr the capacitr wa given a 0 V. Thi reduce the number f unknwn frm tw t ne: C t t ( ) τ 0 ( 0) 0 v t A Be v A Be B A C t t τ τ vc ( t ) A Ae A 1 e The variable A i determined by examining the circuit after a lng time. The capacitr then appear like an pen and the vltage acr it i v R C ( ) V R R R The anwer i (B) ince A v C ( ) : 1. Anwer i (C) The Thevenin equivalent i 1 τ vc ( ) A 1 e A ( 1 0) A K. Kaier 006 Verin 04/13/07 14

15 R th V c while the Nrtn equivalent i I c R th where Ic Vc Rth. (A) i nt true ince the actual circuit generally de nt have the ame net internal reitance a the reitance f it Thevenin and Nrtn equivalent circuit. Althugh (B) eem very reanable, it i generally nt true. Imagine that a lad reitr R i cnnected acr bth equivalent circuit. The pwer diipated by the tw equivalent circuit are V c Vc R PTHEV Rth, PNORT Rth Rth R Rth Rth R Thee pwer are identical nly when R Rth. Thu, if the temperature f tw black bxe, ne cntaining the Thevenin equivalent and ne cntaining the Nrtn equivalent f the ame circuit, are meaured under a given lad, the temperature meaured wuld nt necearily be the ame. Imagine lading bth bxe with a hrt r an pen: V c PTHEV Rth, PNORT 0 Rth V c Vc V c PTHEV Rth 0, PNORT Rth Rth Rth Rth Rth Rth (D) i al nt crrect. Althugh in the literature individual metime attempt t determine a Thevenin equivalent when a device i nnlinear and the utput vltage wing i large (where mall ignal analyi cannt be applied), the equivalent ingle circuit (unle the equivalent element are al a functin f me parameter) i ften f limited value and far frm being an equivalent circuit. The anwer i (C). Frm the utide wrld, by definitin, the lad repne that the equivalent circuit generate when cnnected t a lad i identical t the actual circuit repne. 13. Anwer i (D) (A) i nt crrect. A battery impedance i ften lw. Cnnecting a lwimpedance lad acr the battery culd exceively lad it. The battery vltage culd fall, and thi lad acr the battery culd caue an explin. (B) i nt crrect. Thi matching i nt perfrmed fr maximum pwer tranfer but t btain mre eaily the S parameter. K. Kaier 006 Verin 04/13/07 15

16 (C) i nt crrect. The lad impedance at the end f a ly line i ften elected t be equal t the characteritic impedance f the line t help eliminate peky reflectin. Althugh me prfer might diagree with my lutin t thi prblem, (D) i the bet anwer. 14. Anwer i (D) Thi prblem wa included nt becaue I think the maximum pwer tranfer therem i imprtant that it i wrthy f being teted, but becaue many tet maker eem t lve prblem invlving it (e.g., pleae ee the fficial ample tet prvided by NCEES). When dealing with an ideal tranfrmer withut a plit ecndary, the impedance lking int the primary with a lad impedance f Z i given by Z Zin N N1 where N N 1 i the rati f the turn n the ecndary t the primary cil f the tranfrmer. Sme urce refer t thi rati a n while ther refer t it a 1/n. With a plit ecndary, the input impedance f each eparate lad i tranfrmed t the primary by it repective turn rati; hwever, the tw tranfrmed impedance are nt in erie but in parallel n the primary ide. Thu, the ttal impedance lking int the primary ide f the tranfrmer i Z in 1 1 ( ) ( j ) ( j ) Z j N j j0 N 3 N N j3.8 Ω If the Thévenin r equivalent impedance f the urce i Zth Rth jx th, then maximum pwer i delivered t a lad when Z Rth jx th. The cmplex cnjugate f Z in i given in (D). (If the tranfrmed impedance were incrrectly place in erie, the anwer wuld be 11 j0. The cnjugate f thi incrrect reult i given in (A).) 15. Anwer i (C) Thi i a claical circuit that i uually lved in an intrductry cure in electrmagnetic a a magnetic circuit prblem. Thi magnetic circuit de nt cntain reitance but reluctance. The reluctance f the cre and the air gap are in erie. The reluctance decreae with increaing permeability (a magnetic material prperty). The crrepnding reluctance are π R g g R c, R g µ µ A µ A r K. Kaier 006 Verin 04/13/07 16

17 The ttal reluctance i the um f thee reluctance. A the permeability f the cre increae, the net reluctance decreae. Therefre, (A) i reanable. The flux in the cre and gap, which i analgu t the current in an electric circuit, i NI NI NI µ r µ A Φ R c R π R g g g π R g ( µ r 1 ) µ r µ A µ A where the driving frce i NI, which i analgu t a urce vltage. The magnetic flux denitie in bth regin are Φ NI µ r µ Bg Bc A π R g µ 1 ( ) and the magnetic field intenitie are Bc NI Bg NI µ r Hc, H g µ µ π R g µ 1 µ π R g µ 1 r ( ) ( ) r r r The magnetic field intenitie (nt the magnetic flux denitie) are nt equal. A i clear frm thi exprein, the magnetic field intenitie in bth the cre and air gap increae with I; hence, (D) i crrect. The cre le generally increae with increaing field. The net inductance een by a urce cnnected acr the cil i N Φ N Aµ r µ I π R g µ 1 ( ) A i expected (at leat fr linear material), thi inductance i nt a functin f the current. A the gap length increae, the inductance decreae. Thi i reanable ince inductance tend t increae with the effective permeability f the verall flux path, and the effective path permeability will decreae a the air gap pacing increae. (B) i reanable. A the cr-ectinal area f the cre increae, the inductance increae (imilar t tandard inductr with a imple rd cre). Since the impedance f the abve inductance i given a jω, the current I i given by V ( 1) V V π R g µ r I jω N Aµ r µ jω N Aµ r µ jω π R g µ 1 ( ) A A increae, the current decreae and the pwer factr (PF) a een by the urce decreae. (The pwer factr decreae with increaing inductance and increae with increaing reitance fr a imple R circuit.) Thu, (C) i nt crrect. r r 16. Anwer i (C) The cncept f pwer factr (PF) i very imprtant in pwer engineering. The pwer factr i defined a PF cθ where θ i the angle between the current int and the vltage acr the lad (input impedance lking int the mill in thi prblem). The pwer triangle i helpful in lving prblem f thi type: K. Kaier 006 Verin 04/13/07 17

18 Q cθ 0.7 θ P MW where the cmplex pwer i defined a S P jq VA Uing a little trignmetry, the imaginary pwer i Q tan ( 45.6 ) Q MVAR 6 10 agging pwer factr implie that the paper mill, a een by the external pwer cnnectin, i inductive in nature. Thi i why the Q given i pitive. If the pwer factr wa zer, then the input impedance lking int the mill wuld be purely inductive (r capacitive). If the pwer factr wa ne, it maximum pible value, the input impedance wuld be purely reitive. When a reactive lad, uch a a capacitr, i added in parallel with a lad that require crrectin, the Q add directly. Adding a reactive lad in parallel (nt erie) de nt affect the real pwer delivered t the lad (fr a given urce vltage). Thu, the new Q i Q MVAR 1. MVAR0.8 MVAR new which i till inductive, and the new phae angle i Qnew 0.8 MVAR tan θ new θ new P MW Thi crrepnd t a PF f ( ) increaed). The clet anwer i (C). c The pwer factr i imprved (i.e., 17. Anwer i (B) Fr either a balanced r Y lad, the ttal real pwer delivered t all three lad i given by Pttal 3 Vrm Irm cθ 3 Vrm Irm PF where V rm i the magnitude f the rm line-t-line vltage, I rm i the magnitude f the rm line current, θ i the phae angle between thi vltage and current, and PF i the pwer factr. Thi i an eay prblem t lve if thi relatinhip fr pwer i knwn: 3 Pttal Irm 440 A rm 3 Vrm PF The leading infrmatin i nt needed fr thi prblem. The anwer i (B). Smetime, lad vltage and current are given are requeted. Whether the line-tline vltage and line current are equal t the lad vltage and current, repectively, i dependent n the lad type. It i actually quite eay t determine the magnitude f the K. Kaier 006 Verin 04/13/07 18

19 vltage and current frm the line-t-line vltage and line current and vice vera. Thee relatinhip are given in the fllwing figure. Vl t l Il t l V I Z Z lad V Vl t l I Z I l t l 3 Vl t l Il t l V neutral I Z Z Vl t l V Z 3 I Il t l Y lad Fr a lad, the vltage acr any f the lad i the ame a the line-t-line vltage while the current in any f the lad i 1 3 time the line current. Fr the Y lad, the line current i the ame a the current in any f the lad while the vltage acr any f the lad i 1 3 time the line-t-line vltage. Frequently, unle tated therwie, fr pwer-related prblem, when a vltage i prvided it i prbably the linet-line vltage (and an rm value). 18. Anwer i (B) Fr either a balanced r Y lad, the ttal real pwer delivered t all three lad i given by P 3 V I cθ 3 V I PF ttal rm rm rm rm where V rm i the magnitude f the rm line-t-line vltage, I rm i the magnitude f the rm line current, θ i the phae angle between thi vltage and current, and PF i the pwer factr. Since the vltage at the mtr r lad i given, the impedance f each line leading t the cnductr i nt needed. The 0 HP rating fr the mtr i the deliverable pwer frm the mtr (nt the input pwer). T cnvert HP t watt, multiply by 746: W 0 HP kw HP The actual input pwer t the mtr i a functin f the le r efficiency f the mtr. Thu, the input pwer i 14.9 kw kw, and the line current i (B): 3 Pttal Irm 5 A rm 3 Vrm PF The leading infrmatin i nt needed fr thi prblem. The anwer i (B). 19. Anwer i (A) Since the inductr appear like a hrt at lw frequencie and an pen at high frequencie, thi i a high-pa filter. The cutff frequency fr thi ingle-rder filter can K. Kaier 006 Verin 04/13/07 19

20 be determined tw different way. Fr thi ingle-rder filter, the invere f the time cntant i thi cutff frequency. The time cntant i equal t τ Req R R R R R R Therefre, taking the invere f thi time cntant, it i clear that (B) cannt be the anwer. The cutff frequency can al be determine be writing the exprein fr the vltage gain exprein, uing vltage diviin, and then rewriting the reult in tandard frm fr quick Bde magnitude pltting: jω R V jω R jω R jω R V ( jω R ) R jω R R jω R jω R RR jω R jω R jω R jω ( R R ) RR ω RR j 1 RR R R The cutff frequency i een belw the variable ω, in the denminatr. Althugh (D) ha the crrect cutff frequency, it al cannt be the crrect Bde magnitude plt ince it lpe i 40 db/decade. A firt-rder filter, cntaining nly a ingle inductr, can nly have a repne with a lpe f 0 db/decade. At high frequencie, the inductr appear like an pen and the vltage gain i given by R R R Or, uing the previu exprein and letting the frequency apprach infinity: V j R j R R V j R j R RR j R j R R R The anwer i therefre (A) Anwer i (C) Depending n the handbk() that yu have available, yu may have the Furier erie repreentatin fr thi full-rectified inuidal wavefrm with added dc ffet. If yu have plenty f extra time n the exam, the definitin fr the Furier erie may al be ued: K. Kaier 006 Verin 04/13/07 0

21 a π n π n x( t) an c t bn in t n 1 T T t T π n where an x ( t ) c t dt T T t t T π n bn x ( t ) in t dt T T t0 Hwever, it i nt neceary t ue either a table r thi integral-baed definitin t btain the bet anwer fr thi prblem. Firt, nte that the dc r average value f the wavefrm i between V dc and V dc A. Thi eliminate (A) and (B) ince bth f thee dc value (the nt a functin f frequency) are le than V dc : A 4A 3π 4A 5π 4A 7π Vdc c t c t c t 13 π 3π T 15π T 35π T dc value A 4A π 4A 4π 4A 6π Vdc c t c t c t 13 π 3π T 15π T 35π T dc value Secnd, nte that thi wavefrm i even (i.e., ymmetrical abut the y axi). Thu, it erie huld nly be a functin f cine (nt ine) when thi trignmetric frm f the erie i given (i.e., ne withut phae angle in the argument). Recall that cine i an even functin, x ( t ) x ( t ), while ine i an dd functin, x ( t ) x ( t ). Thu, the mt likely anwer i (C). In thi cae, thi i the erie (becaue I pulled it directly frm my bk) Anwer i (B) Thi i a imple emicnductr prblem. Anwer (C) and (D) are bviuly incrrect (epecially (D)). Since each brn atm accept ne electrn, hle ( pitive charge r area lacking an electrn) are prduced in the ilicn. Brn i an acceptr. Thu, the material i cnidered a p-type emicnductr, (B). 13. Anwer i (C) Strain gauge are extremely imprtant device ued in intrumentatin but are nly ccainally dicued in undergraduate electrical engineering cure. A change in the train n the gauge will generate a mall reitance change, which can be detected with the help f a Wheattne Bridge. T wrk thi prblem, the definitin f gauge factr huld be knwn: R G R Where R and are the change in reitance and length, repectively, and R and are the manufacturer upplied nminal gauge reitance and length, repectively. Subtituting the given infrmatin int the previu exprein K. Kaier 006 Verin 04/13/07 1

22 The anwer i (C) μinche Anwer i (D) Althugh (A) eem reanable, it i nt crrect. By definitin, the renant frequency i the frequency() where the impedance i purely real. The impedance een by the urce i equal t 1 ( R jω ) 1 1 ( ) j ω C Z R R jω R1 jω C 1 R jω jω C ( R jω ) ( R jω ) 1 ω C jω CR R1 R 1 jω CR ω C 1 1 ω C jω CR 1 ω C jω CR 3 R ω RC jω CR jω jω C ω RC R1 1 ω C ω CR 1 ( ) ( ) R ω R C ω R C R ( 1 ω C ) ( ω CR ) ( 1 ω C ) ( ω CR ) j ω CR ω ω C 3 Setting the imaginary term prtin f the impedance equal t zer, the renant frequency i btained: CR 1 R ω CR ω ω C ω ( CR ω C ) ω C C 3 0 The renant frequency, ω, i nt equal t the cmmnly quted reult f 1 C, which i the renant frequency fr imple erie and parallel RC circuit. Mt ther circuit d nt have a renant frequency f 1 C. It wa nt expected that the renant frequency be btained in the abve manner in a few minute. Intead, I wa teting whether yu realized that the renant frequency f mt RC circuit wa nt imply 1 C. (B) i nt crrect. Althugh the maximum r minimum value in the current thrugh r vltage acr me element can be near r at the renant frequency, in general it i nt necearily at the renant frequency. (C) i nt crrect. If the Bde magnitude repne fr the current thrugh and R i determined, it wuld cntain everal break r cutff frequencie. Often when the phrae lw frequency r high frequency are given, the frequency() i relative t ne r mre f thee break frequencie. If the frequency i very high, the inductr appear like an pen and the capacitr appear like a hrt. Thu, the current frm the urce mainly pae thrugh C. K. Kaier 006 Verin 04/13/07

23 If the frequency i very lw, the inductr appear like a hrt and the capacitr appear like an pen. Thu, the current frm the urce mainly pae thrugh and R. Thu, (D) i the anwer Anwer i (B) The v ( t ) veru i ( ) v t plt fr thi limiter/clipper circuit i given belw: v ( t ).65 V i ( ) v t.65 V When the input vltage exceed.65 V, the dide i revered biaed, and the utput vltage i equal t the input vltage (auming the lad acr the utput ha a very high impedance). Fr vltage le than.65 V, uch a 5 V, the dide i frward biaed, and the utput vltage i clipped r limited t.65 V. When the input vltage i a inuid with an amplitude f 5 V, the bttm half f the ine i clipped at a value f.65 V. The anwer i (B) Anwer i (D) Careful! I thi amplifier perating in the active regin? The dc vltage at the emitter i abut V V and the emitter current i e Ve I e kω 3.7 ma Since β i high, thi i abut equal t the cllectr current (when perating in the active regin): β 100 IC I E 3.7 ma β ma The cllectr vltage i then V V C ( ) ( ) Thi reult de nt make any ene, epecially ince all f the applied vltage are greater than zer. The large bae vltage i driving the tranitr int aturatin. Bth the PN junctin f the bae-emitter and the PN junctin f the bae-cllectr are frward biaed. The vltage at the cllectr i lw ince the cllectr current i great. When in aturatin, the tranitr V CESAT 0. V can be ued t determine the vltage at the cllectr: V V V V C CESAT e ( ) The pwer diipated by the cllectr reitr i therefre K. Kaier 006 Verin 04/13/07 3

24 The anwer i (D). P ( 1 7.5) V R k mw 136. Anwer i (B) Since β i mall, d nt aume that I E i apprximately equal t I C. Fr thi PNPbaed circuit, VE 7.4 V, VB 6.7 V, VC 4 V VEB 0.7 V, VCB.7 V Thu, the EB junctin i frward biaed and the CB junctin i revered biaed. Thi tranitr i perating in the active regin and the tandard relatinhip between the emitter, cllectr, and bae current can be ued. Slving fr the cllectr current, VC I C.0 ma kω The emitter current i therefre (B): β I E IC ( ma ).7 ma β 3 Thi current can be checked by uing the bae current VB I B 0.67 ma 10 kω t determine the emitter current: I I β ma 4.7 ma E B ( ) 137. Anwer (C) Grunding i an imprtant tpic in electrical afety (and electrical nie). A typical undergraduate (and graduate) prgram in electrical and cmputer engineering i frequently lacking in thi area. Hpefully, the lutin t thi quetin will help in the undertanding ne apect f afety grunding. (A) i nt crrect (ee the electrical cde when the garage/barn i nt attached t the hue). By cnnecting bth the main panel t a lcal earth grund and the ubpanel t anther lcal earth grund, the tw panel will prbably be at different ptential. Real earth ha a finite cnductivity and thu a ptentially dangeru vltage drp will exit between thee tw panel. Thi i referred t a multiple-pint grunding. (B) i definitely nt crrect. Thi add a third grund fr thi ne cable further cmplicating the grund vltage/current iue. It i bet t refer t the electric cde when plicing a cable. (When i plicing int a cable fr grunding purpe acceptable?) (C) i the crrect lutin. There i ne earth grund at the main panel. Thi i referred t a a ingle-pint grund. (D) i nt crrect. In mt cae when an earth grund i cnveniently available and ha a decent cnductivity, the main panel huld be lcally grunded. There are ituatin where the main panel can be grunded t a (metal) water pipe. Again, ee the electric cde fr detail. A GFCI i nt a ubtitute fr an earth grund Anwer i (B) K. Kaier 006 Verin 04/13/07 4

25 When yu wrked thi prblem, did yu fall int the trap f lving fr the exact value f the input impedance (and wating many minute f valuable teting time) uing the fllwing exprein? 10 Zin j1.45k j0k 14.1 j130k.3m 3m j4. j39k Or, did yu quickly perfrm the analyi in yur head and determine the clet anwer wa (B)? N, I am nt the rain man f circuit analyi. Recall that the impedance f tw parallel element, ne much larger than the ther, i apprximately equal t the maller element: Z1Z Z1Z Z1 Z Z1 when Z1 < < Z Z1 Z Z When the tw element are in erie, the net impedance i apprximately equal t the larger element: Z Z Z when Z < < Z 1 1 Uing thee tw apprximatin, the impedance fr thi circuit can be analyzed in le than a minute withut the ue f a calculatr (r lide rule): 10 Zin j0k 14.1 j130k.3m 3m j4. j39k ( 100) ( { } ) j0k j130k.3m 3m j4. j39k ( ) ( ).3m 3m j4. j39k 5.3m j4. j39k 5.3m j Anwer i (A) Thi i a bridge biplar tranient prtectin netwrk that ue ne zener dide and fur (le expenive) witching dide.. It can handle tranient vltage urge f bth plaritie. With a pitive vltage f 14 V, dide D 3 and D will be frward biaed and D 5 will be revered biaed beynd it zener breakdwn vltage, thu al turned n. The anwer i (A). With a negative vltage f 14 V, dide D 1 and D 4 will be frward biaed and D 5 will be again revered biaed beynd it zener breakdwn vltage and turned n Anwer i (B) The ttal real pwer delivered t the electric heater and ump pump (aumed cnnected in parallel) huld be the ame after the parallel capacitance i added fr pwer factr crrectin. The cmplex pwer delivered t a lad i the um f a real and an imaginary term: Q S P jq VA where tan θ P K. Kaier 006 Verin 04/13/07 5

26 The imaginary pwer abrbed by the heater i 3 1 Qh Ph tan θ h tan c ( 0.98) 84 VAR Since thi pwer i lagging, it i inductive r pitive. The imaginary pwer abrbed by the ump pump i al lagging and given a 3,000 VAR. Thu, the ttal VAR f the tw lad i abut 3.3 kvar. The real pwer aciated with thi ump pump i P 3 Q tan θ tan c ( 0.45) 11.6 kw Thu, the ttal real pwer abrbed by the tw lad i kw. Since thi real pwer de nt change with the additin f the capacitance, the required Q can be determined: 3 1 Qr Pr tan θ r tan c ( 0.95) 4.3 kvar T reduce 3.3 kvar t 4.3 kvar, negative reactance r capacitance i needed with an aciated VAR f kvar. Since the vltage acr thi capacitance i knwn, 10 V rm, the capacitance can be readily determined: Q 3 Vrm Q ω CVrm C 1 π fvrm π ( 60) ( 10) 3.5 mf ω C The anwer i (B). Q θ cθ PF 1 θ c ( PF) P K. Kaier 006 Verin 04/13/07 6