EECE 210 Final, Fall 2013

EECE 210 Final Exam, December 18, 2013 3% (Primal Question) A. The switch is closed at t = 0, with the capacitor initially charged to 5 V in the polarity of vC. Derive the expression for vC as a function of time for t  0. Solution: The initial value of vC is 5 V. It final value is zero.  = RC = 0.5 s. It follows that V. Version 1: VC0 = 5 V, V Version 2: VC0 = 10 V, V Version 3: VC0 = 15 V, V Version 4: VC0 = 20 V, V Version 5: VC0 = 25 V, V. 3% (Primal Question) B. The switch is closed at t = 0, with the inductor initially uncharged. Derive the expression for i as a function of time for t  0, given VSRC = 1 V. Solution: The initial value of i is zero, and the final value is 2 A.  = L/R = 2 s. It follows that A. Version 1: VSRC = 1 V, A Version 2: VSRC = 2 V, A Version 3: VSRC = 3 V, A Version 4: VSRC = 4 V, A Version 5: VSRC = 5 V, A. 3% (Primal Question) C. When the switch is closed at t = 0, the initial voltage across the capacitor is 10 V and the initial current through the inductor is 1 A. Determine the values of vC and iL just after the switch is closed, given R = 2 . Solution: The values of vC and iL remain the same just after switching as just before 1 + – 0.5  1 H i VSRC t = 0 R 1 H iL t = 0 – vC + 0.25 F 0.5 F t = 0 1  + – vC switching. Version 1: VC0 = 10 V, IL0 = 1 A Version 2: VC0 = 15 V, IL0 = 2 A Version 3: VC0 = 20 V, IL0 = 3 A Version 4: VC0 = 25 V, IL0 = 4 A Version 5: VC0 = 30 V, IL0 = 5 A. 3% (Primal Question) D. Specify whether the response for t  0 in the preceding problem is overdamped, critically damped, or underdamped if R = 2 . Solution: rad/s.  = R/2L = R/2. The response is critically damped if R = 4 , is overdamped if R > 4 , and is underdamped if R < 4 . Version 1: R = 2 , response is underdamped Version 2: R = 3 , response is underdamped Version 3: R = 4 , response is critically damped Version 4: R = 5 , response is overdamped Version 5: R = 6 , response is overdamped. 3% 1. Determine Req between terminals ‘ab’, assuming all resistances are 1 . A. 6  B. 3.6 C. 2.4  D. 1.2  E. 4.8  Solution: Two R’s are in parallel and the combination in series with R is 3R/2. In parallel with R, this gives 3R/5. Hence, Req = 6R/5. Version 1: R = 1 , Req = 6/5 = 1.2  Version 2: R = 2 , Req = 12/5 = 2.4  Version 3: R = 3 , Req = 18/5 = 3.6  Version 4: R = 4 , Req = 24/5 = 4.8  Version 5: R = 5 , Req = 30/5 = 6 . 3% 2 a b vL, V 1 2 0 t, s -2 2 4 2. The voltage vL is applied to the initially uncharged inductor. Determine the value of t at which iL = -0.5 A. A. 4 s B. 3.75 s C. 4.25 s D. 3.5 s E. 3.25 s Solution: iL = -K A when the flux linkage  in the inductor reverses and is -K1 = -K Vs. At t = 3 s, the net positive area under the curve is: (1/2)22 – (1/2) 21 = 1 Vs. The time beyond 3 s at which the area becomes -K Vs is (K + 1). This occurs at t = 3 + (K + 1)/2 = 3.5 + K/2 s. Version 1: i = -0.5 A, t = 3.5 + 0.5/2 = 3.75 s Version 2: i = -1 A, t = 3.5 + 1/2 = 4 s Version 3: i = -1.5 A, t = 3.5 + 1.5/2 = 4.25 s Version 4: i = -2 A, t = 3.5 + 2/2 = 4.5 s Version 5: i = -2.5 A, t = 3.5 + 2.5/2 = 4.75 s. 3% 3. Determine the energy stored in the circuit under steady-state conditions, given VSRC = 1 V (Hint: consider Leq for energy calculation). A. 14 J B. 10.5 J C. 1.75 J D. 3.5 J E. 7 J Solution: Leq = 2 + 4 – 2 = 4 H. The circuit under steady- state conditions is a shown, with inductors acting as short circuits and capacitors as open circuits. The voltage across the capacitors is VSRC/2, and the energy stored is J. The energy stored in the 2 H inductor is = J. The energy stored in the 4 H inductor is 3 4 H 1 H 2 H + – VSRC 1 F 1 F 1  1  2 H 0.5  + – vL iL 1 H 4 H + – VSRC 1 F 1 F 1  1  2 H 0.5  VSRC VSRC/2 + – VSRC/2 = J. The total energy stored is . Version 1: VSRC = 1 V, W = = 7/4 = 1.75 J Version 2: VSRC = 2 V, W = = 74/4 = 7 J Version 3: VSRC = 3 V, W = = 79/4 = 15.75 J Version 4: VSRC = 4 V, W = = 716/4 = 28 J Version 5: VSRC = 5 V, W = = 725/4 = 43.75 J. 3% 4. Determine Zeq between terminals ‘ab’, given all capacitances are 1 F, all inductors are 1 H, and  = 1 rad/s. A. 0 B. j2.75  C. j4.89  D. -j2.75  E. -j4.89  Solution: Leq = 1 + 1/2 +1/3 = 11/6 H; Ceq is 1 F in series with 2 F, in series with 3F. The last two are 6/5 F, and Ceq = 6/11 F. Zeq = j(11/6 – 11/6). Version 1:  = 1 rad/s; Zeq = j(11/6 – 11/6) = 0 Version 2:  = 2 rad/s; Zeq = j(112/6 – 11/62) = j2.75  Version 3:  = 3 rad/s; Zeq = j(113/6 – 11/63) = j4.89  Version 4:  = 4 rad/s; Zeq = j(114/6 – 11/64) = j6.88  Version 5:  = 5 rad/s; Zeq = j(115/6 – 11/65) = j8.8 . 3% 5. Determine iC as a phasor, given vSRC = sin2t V, and assuming cos2t to have zero phase angle. A. 1.41-45 V B. 0.70790 V C. 0.707+45 V D. 1.4145 V E. 0.707-45 V Solution: vSRC = Vmsin2t = Vmcos(2t – 90) V;  = 2 rad/s, so that the circuit in the frequency domain is as 4 – + vSRC 1 H 0.25 F ix 0.5dix/dt iC  – + Vm-90 j2  Ix jIx IC  -j2  a b shown. From KCL, IC = (1 + j)Ix. From KVL around the LHS mesh, Vm-90 = j2Ix – j2(1+ j)Ix = 2Ix = 2IC/(1 + j). Hence, IC = 0.5(1 + j)(Vm-90) = 0.5 Vm-45 V. Version 1: Vm = 1 V, IC = 0.5 (1)-45 V = 0.707-45 V Version 2: Vm = 2 V, IC = 0.5 (2)-45 V = 1.41-45 V Version 3: Vm = 3 V, IC = 0.5 (3)-45 V = 2.12-45 V Version 4: Vm = 4 V, IC = 0.5 (4)-45 V = 2.83-45 V Version 5: Vm = 5 V, IC = 0.5 (5)-45 V = 3.54-45 V. 3% 6. Determine Norton’s current looking into terminals ‘ab’ (not including RL), given Vm = 1 V. A. 7.0745 A B. 3.54-45 A C. 3.540 A D. 3.5445 A E. 7.070 A Solution: The independent current source and the 1  resistor in series with the dependent source are redundant as far as IN is concerned. The circuit becomes a shown. IN = Ix; the parallel impedance of L and C is; (-j2)(j)/(-j2 + j) = j2 . From KVL, Vm45 = (2 + j2)Ix. It follows that IN = Ix = Vm45/2(1 + j) = A. Version 1: Vm = 10 V, IN = = 3.540 A Version 2: Vm = 20 V, IN = = 7.070 A Version 3: Vm = 30 V, IN = = 10.60 A Version 4: Vm = 40 V, IN = = 14.10 A Version 5: Vm = 50 V, IN = = 17.70 A. 3% 5 – +  10 A Vm45 -j2  j  Ix 1  1  a b RL Ix – + Vm45 -j2  j  Ix 1  a b Ix IN 7. Determine ZTh looking into terminals ‘ab’, given Z = j9 . A. Infinite B. j4  C. -j2.5  D. -j4  E. j8  Solution: Applying a test source VT, with the current source replaced by an open circuit, it is seen that IT = 2.25VT/Z, which gives ZTh = 4Z/9 . Version 1: Z = j9 , ZTh = j4  Version 2: Z = -j9 , ZTh = -j4  Version 3: Z = j18 , ZTh = j8  Version 4: Z = -j18, ZTh = -j8  Version 5: Z = j4.5 , ZTh = j2 . 3% 8. The capacitor is charged to 10 V at the instant the switch is closed at t = 0. Determine the damping coefficient  of the circuit for t > 0, given R = 1 . A. 1 rad/s B. 1.5 rad/s C. 0.125 rad/s D. 0.167 rad/s E. 0.5 rad/s Solution: Using the T-equivalent circuit, (-1)2/(2 – 1) = -2 H. In series with 3 H this gives 1 H. Hence,  = R/2L = R/2 rad/s. Version 1: R = 1 ,  = R/2 = 0.5 rad/s Version 2: R = 2 ,  = R/2 = 1 rad/s Version 3: R = 3 ,  = R/2 = 1.5 rad/s Version 4: R = 4 ,  = R/2 = 2 rad/s Version 5: R = 5 ,  = Ri2 = 2.5 rad/s. 3% 9. The switch is opened at t = 0 after being closed for a long time. Determine iC just 6 2 H 1 H 1 H 1 F R t = 0 2 H 1 F R 3 H -1 H 1:2  1 A Z a b + – iC VSRC 1 k 1 k 0.5 F 5 mH t = 0 1:2 Z – + VT 0.5VT + – IT Z VT 75 0. Z VT 1.5 after the switch is opened, given VSRC = 1 V. A. 0.5 mA B. 0 C. 2 mA D. 1.5 mA E. 1 mA Solution: Before the switch is opened, iL = VSRC/2 mA, directed downward. After the switch is opened, iC = iL = 0.5 mA. Version 1: VSRC = 1 V, iC = 0.5 mA Version 2: VSRC = 2 V, iC = 1 mA Version 3: VSRC = 3 V, iC = 1.5 mA Version 4: VSRC = 4 V, iC = 2 mA Version 5: VSRC = 5 V, iC = 2.5 mA. 3% 10. The circuit is in a steady state, with i = sin0t A. Determine the smallest t > 0 at which the instantaneous energy stored in the inductor is equal to that stored in the capacitor, given L = 1 H. A. 1.57 s B. 1.76 s C. 3.51 s D. 0.79 s E. 1.36 s Solution: If i = Imsin0t A, v = Ldi/dt = 0LImcos0t V. When the two instantaneous energies are equal, , which gives sin0t = cos0t, or tan0t = 1, or 0t = /4, or t = /40 = = s. Version 1: L = 1 H, s Version 2: L = 2 H, s Version 3: L = 3 H, s Version 4: L = 4 H, s Version 5: L = 5 H, s. 3% 11. The switch is closed at t = 0, after being open for a long time. It is 7 L i 1 F R C 1 A t = 0 1 H kR iL  – + 4(1 + j) A Ix 5Ix -j5  j40  j15  j10  10  -j40  a b found that iL(t) = A. Determine R, given k = 1. A. 0.6  B. 1  C. 0.667  D. 0.75  E. 0.625  Solution: 0 = 1 rad/s. since the response is critically damped. C = = 1 F. ; Rp = 1/2 = kR/(1 + k); R = (1 + k)/2k . Version 1: k = 1; R = (1 + k)/2k = 2/2 = 1  Version 2: k = 2; R = (1 + k)/2k = 3/4 = 0.75  Version 3: k = 3; R = (1 + k)/2k = 4/6 = 0.667  Version 4: k = 4; R = (1 + k)/2k = 5/8 = 0.625  Version 5: k = 5; R = (1 + k)/2k = 6/10 = 0.6 . 3% 12. The switch is opened at t = 0, after being closed for a long time. Determine dv/dt just after the switch opens, given ISRC = 1 mA. A. -1 V/ms B. -2 V/ms C. 0 D. 1 V/ms E. 2V/ms Solution: Before the switch opens, v = 0 and iL = 0. Both remain zero just after the switch is opened. ISRC will therefore initially pass through the capacitor, which makes ISRC = -Cdv/dt, It follows that dv/dt = -ISRC/C = -110-3/110-6 = -103 V/s = -1 V/ms. Version 1: ISRC = 1 mA, dv/dt = -1 V/ms Version 2: ISRC = 2 mA, dv/dt = -2 V/ms Version 3: ISRC = 3 mA, dv/dt = -3 V/ms Version 4: ISRC = 4 mA, dv/dt = -4 V/ms Version 5: ISRC = 5 mA, dv/dt = -5 V/ms. 13% 13. Derive TEC looking into 8 0.1 S + – v 1 mH ISRC t = 0 1 F  – + ISRC Ix 5Ix j30  10  -j40  a b j5  -j5  j10  5Ix I = 0 ISRC – Ix terminal ‘ab’ (7 grades for VTh, 6 grades for ZTh). Solution: Replacing the linear transformer by its T-equivalent circuit, it is seen that on open circuit, the -j40  in series with j10  gives -j30 . In parallel with j30 , this results in an open circuit. The current through the -j5  is zero, so that both terminals of this inductor are at a voltage 5IX. The current in the -j5  impedance is ISRC – Ix. From KVL, 10Ix + j5(ISRC – Ix) = 5Ix. This gives A, and 5IX = 20 V. It follows that VTh = Vab = = 26.67 V. On short circuit, the voltage of the dependent source is still 20 V. The parallel impedance of j10  and j30  is j7.5 . The current through the j5  impedance is 20/(j12.5) A, From current division, ISC is this current multiplied by 3/4. This gives ISC = -j1.2 A. It follows that ZTh = (80/3)/(-j1.2) = j200/9 = j22.22 . 13% 14. Determine Zin looking into terminals ‘ab’. Solution: Let a test voltage of 5VT be applied. The voltages across the two windings are 4VT and VT. w The current in the RC branch is . The current in the inductive branch is: = 0.25VT(1 – j). Ix = 0.4VT(1 + j) – 0.25VT(1 – j) = VT(0.15 + j0.65). From the mmf equation, 4I1 + I1 + VT(0.15 + j0.65) = 0, which gives I1 = -VT(0.03 + j0.13); I = 0.4VT(1 + j) – VT(0.03 + j0.13) = VT(0.37 + j0.27); Zin = 5VT/VT(0.37 + j0.27) = 10.92-36.12 = 8.818 – j6.435 . 9 j30  a b j5  j10  20 V ISC – + j2  1600 turns 400 turns a b 10  -j10  2  – + – + – + Ix I1 Ix + I1 I VT 4VT 5VT j2  1600 turns 400 turns a b Zin 10  -j10  2  13% 15 The switch closes and opens repeatedly at regular intervals, resulting in a periodic voltage v that alternates between an upper level of 4 V and a lower level of 2 V. (a) Derive expressions for v as a function of time for the interval 0  t  t1 and the interval t1  t  t2 (4 grades each); (b) determine t1 and t2 (2 grades and 3 grades). Solution: (a) For 0  t  t1,  = 0.5 ms, and , t is in ms; for t1  t  t2,  = 1 ms, and , t is in ms. (b) For 0  t  t1, , , t1 = 0.5ln2 = 0.35 s. For t1  t  t2, ; t2 = ln2 + 0.5ln2 = 1.5ln2 = 1.04 s. 13% 16. The switch is moved at t = 0 from position ‘a’ to position ‘b’, after being in position ‘a’ for a long time. Determine: iO and vO as functions of time for t  0 (5 grades and 6 grades). Solution: Leq = 0.2 + 0.4 – 0.2 = 0.4 H. The circuit for t  0 is as shown, where IO0 = 12 A; Ix0 = 20 A; VO0 = 24 – 80 = -56 V. In the steady state, VOS = 0, and 2IOS – 4(IOS + 8) = 0, so that IOS = -16 A. The resistance seen by the inductor is obtained by applying a test source, with the independent soirce set to zero. VT + 2IT = 4IT, which gives Reff = 2 , and R/L = 5 s-1. It follows that: A, and V. As a check, LdiO/dt = 0.4(28)(-5) = = vO. 10 + – 1 k 1 k 1 F 12 V – + v 2 V t t1 t2 4 V 0 v 0.2 H + – 4  2  + – vO iO 24 V t = 0 8 A a b – + 0.4 H 0.1 H 2iO 4  + – vO iO 8 A – + 0.4 H 2iO ix 4  IT – + 2IT + – VT