EECE 210 Exam 2, Spring 2016

American University of Beirut Department of Electrical and Computer Engineering EECE 2I0 - Electric Circuits Qutz 2 Closed Book - No Programmable Calculators - 90 minutes April 17,2015 Name: KFY- So Lt/noM Solve the following problems; Provide your answers on the attached Scantron card; This question sheet must be returned with the Scantron card; There is No penalty; Mark with a pencil your LAST NAME, your First name Initial (FI) and your Middle name Initial (MI); Mark your AUB ID Number in the box titled "ID NUMBER"; Write the name of your course instructor on the Scantron card; Use a pencil for marking your answers ; When using an eraser, make sure that you have erased well. This exam has 9 pages. Course Instructors: L. Hamandi, S. Khaddaj, Y. Nasser, and R. Jabr. ID: 1. In the circuit shown in Fig. 1,15: I A and each of the resistances is equal to 5 0. Findfu. 5V (a) 0.25 A @o.sa (c) 0.75 A (d) 1A (e) None of the above I 5 Y rs 3A (a) 10V (b) 20 v (c) 30 V @+ov (e) None of the above 512 30v 2oT7 Jy -3o *5+6 = O lo = o,5A 2o ) -vv\r----_-l I I I l.,{ c ) I 6 I A TX Fig. 1: Circuit for Problem 1 2. lf R: 30 O in the circuit shown in Fig.2, flnd the Th6venin voltage between terminals a,b. \4r.,.= -ffi5, +60 = ?oV + R 6oV 3o-Q 2A 5fr Fig. 2: Circuit for Problem 2 5A- 2t). T 3. If 1s : 5 A in the circuit shown in Fig. 3, find the Th6venin voltage between terminals a,b. (fra v (b) 12 v (c) 14 V (d) 16v 2Vx (e) None of the above T= 5o-lo = ?A lo+z+8 Vx = 2Z= 4V lt Vrp, - -?Vx +/o+ 8J --2(4)+tD+g(2) : 18 V 15 5A Fig. 3: Circuit of Problem 3 In the circuit shown in Fig. 4, Ra:3 O. Find the value of Rz, that results in maximum power being transferred to R1. (@', n (b) 16 O (c) 17 A (d) 18 O (e) None of the above 6r} Fig. 4: Circuit of Problem 4 Kt=&n= to+3+3116 = lS +Z = (5{L 10a 5. If V5 : 120 V in the circuit shown in Fig. 5, find the maximum power that can be delivered to R2. (a) 329.t4w (b) 302.29w (c) 228.57 W @eo.rr* (e) None of the above VS vrn = 4o (p") ( 8x2o + 40) = -lott V 4o+lo - ?8{2 = q6,5+ w 10 g) -vltv1d----! MI wi FI +*c) ffi YYVY M *u @ 40v t?ov Fig. 5: Circuit of Problem 5 6. In the circuit shown in Fig. 6,V7:120 V. Find the component of 15 resulting from the yr voltage source acting alone, with all the other voltage and current sources deactivated. (a) 1A (b) 2A (c) 3A @oo (e) None of the above l2o 5tro +/5 = tlA Fig. 6: Circuit of Problem 6 Tl Jg 7. In the circuit shown in Fig. 7, Vcc:2.5 V. Find the maximum value of Rx that can be used such that the op-amp operates in its linear region. (a)r5Q o'al+ $u 6fuoo .u- (c)45o l+ 2Vr1 +\ (d) 60 o (e) Noneof theabove VD = 3 Vn + I Vn-Vo = c2 'tM V, =o 4,5 =) Vn - o.5 o., Fx + tS -- Rx r,) By+3 o =Ex €4 = 3oc2 .+ T, Fig. 7: Circuit of Problem 7 Each op-amp shown in Fig. 8 is operating in the linear region, and each of the resistances is equal to 10 O. If ys : 6 V and Vr :24 V, find the current 16r. (a) 0.5 A @o.oo (c) 0.7 A (d) 0.8 A (e) None of the above l3r.+6 +Z+ lo 3oIo = 18 I:,I--ZU = O /o -) = o'64 Fig. 8: Circuit of Problem 8 9. The op-amp shown in Fig. 9 is operating in the linear region. R: : 4 kO, and Rr :10 kO. Find [. (a) 1.75 mA (b) 2.375 mA @''"o (d) 4.25 mA (e) None of the above R1 UK Vsnc:1 Y Rr :4 kf), Rz :40 kO, + (t+ ?yu = ttv : 3-A I --ll q? Fig. 9: Circuit of Problem 9 10. If C: ll ruF in the network shown in Fig. 10, find the equivalent capacitance with respect to the termir,als q, b. (a) 0.8 triF (b) 2 ttF @zl pr (d) 3 pF (e) None of the above + tv Iv ,ft lt*t = 6f F 2 C"b !*t 6 3+z 4l,-il 2o =) Cro = 2'47f Fig. l0: Circuit of Problem 10 11. In the circuit shown in Fig. ll, L1 :4H, lrz- 4H, and M:2H. Write the mesh-current equation around mesh 2 (corresponding to mesh-current i2). (a) 6(d\ I dt) - t3(di2l dt) - t4i2 : Q ((b))6(di I ldt\ - t2(dizldt) - t4i2: Q \-7 (c) 6(di1 I dt) - tl(di2l dt) - t4i2 : s (d) 6(dit I dt) - t}(diz/ dt) - t4i, : o (e) None of the above 4H L2 Bn 4n 4 oluz '-+ d+ -6d[, dt zd-(ir-t,) + /Z di, + ltt dt Fig. 1 1: Circuit of Problem 1 I + tqt:z + 4d (tz-\) 4t + 2 olt', E =o iz -0 12. In the circuit shown in Fig. l2,Vs:75 V. The switch has been closed for a long time, and is opened at t = 0. Find the total energy dissipated in the 5 f,) resistor. @0, r ?Lb-) -- 7; (b) t.z.r I 5+ f B (c) 2.7, ?tft) = 3 e- 75t ' (d) 4.8 J (e) None or rhe above 'Prn f4 = S irlO = 7-5 3A 4e ,t5, = O'3{ -tbt 1&- W Jb-',*b, 2.5 r} VS 75v Fig. 12: Circuitof Problem 12 13. In the circuit shown in Fig. 13, the initial currents in inductors L1 and Lz 00 A upward and 4 A downward) have been established by sources not shown. The switch is opened at t : O. Find the value oti2(t)atl:0.82s. l(fJ -_ 6 e-o.St ^ =) tr1) = /?€_ o.St V (a) -5.00 A (b) -4.80 A (c) -a.60 A @ -+.+o e, (e) None of the above Wsn = t={ 0.1 fi lir"(O I ir(,.s2) = -* If;T-".,fu - q o = _tz /;o.0,-^l_/ = eo.Go.S) r" t(4 = _o.vt t.? @ -,)- ,/ - 4, tlo A zil 4^l tr(5H) I f r'tt) I"2 (20 H) firt 4 n ir'^ Fig. 13: Circuit of Problem 13 in the time interval 0<=t<infinity. 14. In the circuit shown in Fig. 14, Vs : 180 V. The two switches operate simultaneously. Switch I (SW1) has been in position a and switch 2 (SW2) has been in position c for a long time. At / = 0, SWI moves to position b ard SW2 moves to position d. Find %(0+). =50 tgo = tioV (a) 10V (b) 20 v (t}ro, (d) 40 v (e) None of the above 6,b1 = /r(t) Vobl =,# 5ollo tso = 3ov b+ lmF r *:o srr/l SW1, 40 ko frtU r0 kfi I v"(r) 15. Fig. 14; Circuitof Problem 14 In the circuit shown in Fig. 15, the initial capacitor voltages are u1(0) : 10 V and u2(0) :5 V. Find thevalueorv2(t)as/-+€. T = 4xlt6, ltlot_= ltto3s ._3 _2rof @rt /(r/ = (to-s)g-z,o' V => i(t) =5r/o e -'- A (b) 13 v (c) r7v 6t^1 = t f*srti'e-'*1* + i (d)2rv 5x65J .-3 (e)Noneofrheabove o = 5l=lor- , (O-,) +E r tf(q S r ttr' (-su) I fr =qv , I 1kCI Vtt+l ' __l_ f l_ v,(t) - z0pF vr(t) - spf Fig. 15: Circuit of Problem 15