2 Problems on Power Circuits and Electromechanics with Solution - Exam | ECE 430, Exams of Electrical and Electronics Engineering

Download 2 Problems on Power Circuits and Electromechanics with Solution - Exam | ECE 430 and more Exams Electrical and Electronics Engineering in PDF only on Docsity! ECE430 Le Spring 2006 2: Exam 1 3: March 1, 2006 4 Name SOLUTION Total: Section (C for Kimball MWF, F for Tate TR) Equations: = I - «, VV Q=— 16 = VI = | z HA Sy = SFT =V3V, 1,40 My = 40x10" yy By, =V3V1, 0080 A=NO=Li 2 0, = V3V, I, sind L=No=8 bf = cos (<7 - <I) mmf (source) = Ni @ > 0 lagging, 8 < 0 > leading mmf (drop) = 24 2 a 92 Peag=8 Dd mmf = 0 around loop 1 aa fon X.=-aa G Hed! = [J-ida %, = 0b Gindl =-£ [Benda ot wye, abe sequence :V, =V, =F, (V3 430° )7= _ fo GBeAda=0 delta, abe sequence : Vi=V,0, = i, (34 “so M LL, waa dt Rio LjnaM Rp Eg Main Transformer Approximate Equivalent Circuit Transformer Equivalent Circuit Problem 1 (25 points) A single-phase source, generating a voltage v(/) = 120V2 cos(377t) , is connected to a single load. The power factor of the load is measured to be 0.8 lagging. A 8000 VAR capacitor is then added in parallel to the load, and the power factor of the load and capacitor combination is measured to be 0.9 leading. i® [} vt) Load 1 g000 vaR L Tk a) Find the average (real) power (P) and reactive power (Q) of the initial load (10 pts) b) Determine the complex impedance (Z,,,, ) of the initial load (5 pts) c) Determine i(¢) both before and after the capacitor is added to the system (10 pts) a) phe orig = 0-8 lag =? #Sorig = 36.87? Sorig > Ports + 5 Qeig ph wf ange OF lead => x By w 25.84" Seap * Porig t j Seay . ig Scap = : - 000 Aan (36. BT) = os tan (> 25:24") = Ss Scap = forig ty (Qoris 0d) Povig oi § we 908. donCreser)= Seria - S208 ~ joncsee7”) ee oo Tenis Jan (36.27) ~ tial- 25-64") Poet Porig orig = btBl ww Borg » Porta Ctan C36,27°9) a = 4861 ve OT oF 120 = ve y= l2a 20 Z toad * b) Fae ze eat ~ j9S6 | orig io Fo = Son ed Fin HFG! = 675 2-56.87 ao Leh) = ¥zU-67 S cos (2774 - 36, 7) ) tvy? 25 = SES ) “ | £ . 3 Seb, (ous jis jFOON” od 84) FFT _ go eas i@t® = & a 7 ° e ese v* 1202.0° 12020 (4) > fz 60- cos (377t +25,84 ‘ wot C) Wa, ~ bX, - 64-4) CR,* B24 Prew Ft 2, - (4,-4.) Caveat O @ Pia, = (Xr Pe) 4 ~ (RB) & bra (Rar Vs) e a (SB ¥s) 4. Be Bn By 4 Rap Pe) a - th } Re “es ALR Ve YO | b “ Jo ne ye) jaye Us Pa ee . > cat) . Gout ~-sLAK i va w\S | Lay | SURAR 6 rseR ies, A? C2S bub a - SL AWK AL ~ bp bay & Ale ShRaR: A, ay l= 62S.buk a) lyF b. 256m ft & Me chim &D ee ee: pare (A) Problem 4 A load is being supplied by a single-phase source (represented here as an ideal voltage source behind a series impedance of 10 + j5 ©) and transformer, as shown in the following diagram. The load, connected to the LV side of the transformer, has a complex impedance of 2.5 +j1.25 Q. 2.54+j1.252 a) Assuming the transformer is ideal, determine the turns ratio that maximizes the average power delivered to the load (7 pts) After performing a short-circuit test, it is determined that the transformer is not ideal, but instead has parameters Rjeq = 1, Xteq = 0.5 Q. The open-circuit test is not performed, therefore Rc and Xp are neglected. Using these parameters, and the turns ratio calculated in part (a): b) Ifthe load is to be supplied at 120 V, find the RMS magnitude of the current leaving the source. (7 pts) c) Ifthe load is to be supplied at 120 V, what is the required RMS voltage magnitude |V,) of the ideal voltage source? (11 pts) _ oy a) 1Bl= at lech [Bole Vwres® “= wig Feil Yas*tast = 2.78 Bn \ = u = . “aan . = of Ens Lee 2 B82 BANE 14,2 -jUG Viz t2eo4 BT. z L Z,S4j\h2S [Eafe }n2 jae [RATA S (ors) Es 7 C1405) I, + 24e20° Zs=!f2-jT% j . (oj SUG? ~j%b)+ CH HSC In2 -y 26) + 2Y0L0 Ves 504 Loov ivel = [504 vl ec) Vs = wv it