EECE 210 Final With Solution, Fall 2014

+ – 10  0.2 F VSRC + – vC t = 0 EECE 210 Final Exam, December 13, 2014 3% A. Two coils are wound on a high-permeability toroidal core of 2 cm2 cross-sectional area, with coil 2 open-circuited. If the magnetic flux density B in the core, in the clockwise sense, due to current in coil 1, is decreasing at a constant rate of 0.1 Wb/cm2/s, determine the magnitude and sign of v2, assuming coil 2 has 10 turns. Solution: 21 = BAN2; . If B is decreasing with time, and current is allowed to flow in coil 2 by connecting a resistor across coil 2 terminals, current will have to enter at the upper terminal of coil 2 to produce a flux aiding that in the core. It follows that the sign of v2 is negative. Version 1: N2 = 10 turns, v2 = -0.2N2 = -2 V Version 2: N2 = 15 turns, v2 = -0.2N2 = -3 V Version 3: N2 = 20 turns, v2 = -0.2N2 = -4 V Version 4: N2 = 25 turns, v2 = -0.2N2 = -5 V Version 5: N2 = 30 turns, v2 = -0.2N2 = -6 V 3% B. The switch is closed at t = 0. Determine vC(t), t  0+, assuming an initial value VC0 = 5 V and VSRC = 10 V. Solution: VCF = VSRC,  = 2 s, so that V, t  0+, t is in s. Version 1: VSRC = 10 V, V, t  0+, t is in s Version 2: VSRC = 15 V, V, t  0+, t is in s Version 3: VSRC = 20 V, V, t  0+, t is in s Version 4: VSRC = 25 V, V, t  0+, t is in s Version 5: VSRC = 30 V, V, t  0+, t is in s 3% C. The switch is closed at t = 0, with an initial voltage VC0 = 5 V, and no energy stored in the inductor. Determine v(0+), iL(0+), and iC(0+), assuming R = 2 . Solution: v(0+) = Vc0, iL(0+) = 0, iC(0+) = -Vc0/R. p1/8 – + vSRC1 Coil 1 Coil 2 R v2 + _ + – 10 mH 10 mF v iL iC t = 0 + – VC0 R  Version 1: VC0 = 5 V; v(0+) = 5 V, iL(0+) = 0, iC(0+) = -Vc0/R = -5/2 = -2.5 A Version 2: VC0 = 10 V; v(0+) = 10 V, iL(0+) = 0, iC(0+) = -Vc0/R = -10/2 = -5 A Version 3: VC0 = 15 V; v(0+) = 15 V, iL(0+) = 0, iC(0+) = -Vc0/R = -15/2 = -7.5 A Version 4: VC0 = 20 V; v(0+) = 20 V, iL(0+) = 0, iC(0+) = -Vc0/R = -20/2 = -10 A Version 5: VC0 = 25 V; v(0+) = 25 V, iL(0+) = 0, iC(0+) = -Vc0/R = -25/2 = -12.5 A 3% D. Specify whether the response for t  0+ in the preceding problem is overdamped, critically damped, or underdamped, assuming R = 0.2 . Solution: rad/s; Version 1: R = 0.2 ; p = 50/0.2 = 250 rad/s, response is overdamped Version 2: R = 0.4 ; p = 50/0.4 = 125 rad/s, response is overdamped Version 3: R = 0.5 ; p = 50/0.5 = 100 rad/s, response is critically damped Version 4: R = 0.8 ; p = 50/0.8 = 62.5 rad/s, response is underdamped Version 5: R = 1 ; p = 50/1 = 50 rad/s, response is underdamped 3% 1. Determine Va assuming all resistances are 1  and the current sources are 1 A each. Solution: From symmetry, each half circuit is as shown. The current in the 2  resistor is I/4 A, and Va = I/2 A. Version 1: I = 1 A, Va = 0.5 V Version 2: I = 2 A, Va = 1 V Version 3: I = 3 A, Va = 1.5 V Version 4: I = 4 A, Va = 2 V Version 5: I = 5 A, Va = 2.5 V 3% 2. Determine Norton’s current IN and Norton’s admittance YN, looking into terminals ‘ab’, assuming VSRC = 590 V. Solution: When terminals ‘ab’ are short-circuited, IN = IX = VSRC/j5 A. When terminals ‘ab’ are open circuited, VTH = -j0.5IX = -j0.5VSRC/j5; when a test source is applied, with VSRC = 0, then IX = 0, so that the independent source behaves as an open circuit, and YN = 1/(-j5) = 2 S. Version 1: VSRC = 590 V, IN = VSRC/j5 = 10 A, YN = j2 S p2/8 – + VSRC j5  -j0.5  IX IX a b Va Va I 1  1  2  Version 2: VSRC = 1090 V, IN = VSRC/j5 = 20 A, YN = j2 S Version 3: VSRC = 1590 V, IN = VSRC/j5 = 30 A, YN = j2 S Version 4: VSRC = 2090 V, IN = VSRC/j5 = 40 A, YN = j2 S Version 5: VSRC = 2590 V, IN = VSRC/j5 = 50 A, YN = j2 S 3% 3. Determine the net average power delivered in the circuit, assuming VSRC = 100 V. Solution: The net average power delivered is equal to the average power dissipated in the 4  resistor. When the current source is set to zero, the current in the inductive branch is A. When the voltage source is set to zero, the current in the inductive branch is A. By superposition, the total current in the inductive branch is . The magnitude of this current is , and the average power dissipated is W. Version 1: VSRC = 100 V, = 2 W Version 2: VSRC = 150 V, = 8 W Version 3: VSRC = 200 V, = 18 W Version 4: VSRC = 250 V, = 32 W Version 5: VSRC = 300 V, = 50 W 3% 4. Determine I so that Vo = 0, assuming that X = 2 . Solution: When Vo = 0, the circuit becomes as shown, with Va = Vb. It follows that j4 = jXI, so that I = (4/X)0 A. Version 1: X = 1 , I = 4/X = 40 A Version 2: X = 2 , I = 4/X = 20 A Version 3: X = 4 , I = 4/X = 10 A Version 4: X = 8 , I = 4/X = 0.50 A Version 5: X = 10 , I = 4/X = 0.40 A 3% 5. Determine iS(t), assuming vSRC(t) = 4cos100t V. Solution: jL = j10010-2 = j ; -j/C = - j/(10010-2) = -j. It follows that the impedance of the paralleled RLC elements is 2 ; iS(t) = vSRC(t)/2 = 2cos100t A p3/8   – + Vo + _ -j5  jX 4  -j4  j2  2Vo I 190 A   jX 4  -j4  j2  I 190 A a b – + vSRC 10 mH 10 mF 2  iS – + VSRC  5-90 A -j  j4  4  Version 1: vSRC(t) = 4cos100t V; iS(t) = vSRC(t)/2 = 2cos100t A Version 2: vSRC(t) = 6cos100t V; iS(t) = vSRC(t)/2 = 3cos100t A Version 3: vSRC(t) = 8cos100t V; iS(t) = vSRC(t)/2 = 4cos100t A Version 4: vSRC(t) = 10cos100t V; iS(t) = vSRC(t)/2 = 5cos100t A Version 5: vSRC(t) = 12cos100t V; iS(t) = vSRC(t)/2 = 6cos100t A 3% 6. Determine the maximum magnitude of vL and its phase angle at this magnitude, given vSRC = 2cost V, and  assuming any value between 0 and . Solution: Maximum current occurs at rad/s and is A. It follows that = V. Version 1: vSRC = 2cost, |vL|max = 2 V, L = 90 Version 2: vSRC = 4cost, |vL|max = 4 V, L = 90 Version 3: vSRC = 6cost, |vL|max = 6 V, L = 90 Version 4: vSRC = 8cost, |vL|max = 8 V, L = 90 Version 5: vSRC = 10cost, |vL|max = 10 V, L = 90 3% 7. Determine Zin, assuming  = 10 rad/s and C = 20 mF. Solution: Leq for the two coupled coils is (2 + 2 – 2) = 2 H; in parallel with 2 H, this gives 1 H, or j10 . The two capacitors in series are C/2, having an impedance -j0.2/C , when C is in F. It follows that Zin = j10 – j0.2/C . Version 1: C = 20 mF, Zin = j10 – j0.2/0.02 = j10 – j10 = 0 Version 2: C = 10 mF, Zin = j10 – j0.2/0.01 = j10 – j20 =-j10  Version 3: C = 5 mF, Zin = j10 – j0.2/0.005 = j10 – j40 =-j30  Version 4: C = 4 mF, Zin = j10 – j0.2/0.004 = j10 – j50 =-j40  Version 5: C = 2 mF, Zin = j10 – j0.2/0.002 = j10 – j100 =-j90  3% 8. Determine Zin, assuming X = 2 . Solution: With a test source applied, the voltage across the autotransformer is V1 – V1 = 0. It follows that VT/IT = Zin = -jX . Version 1: X = 2 , Zin = -jX = -j2  Version 2: X = 4 , Zin = -jX = -j4  Version 3: X = 6 , Zin = -jX = -j6  Version 4: X = 8 , Zin = -jX = -j8  p4/8 100 turns j10  100 turns Zin -jX 100 turns 100 -jX V1 + _ _ – + VT IT – + vSRC 1 mH 0.1 F 100  vL + _ 2 H 2 H 2 H 1 H C C Zin 0.1 F 5  ISRC 5  2.5 H + – vS vL + _ vC + _ Version 5: X = 10 , Zin = -jX = -j10  3% 9. Determine the total energy delivered by the battery from t = 0, when the switch is closed, till t  , assuming VSRC = 10 V and no initial charge on the capacitor. Solution: The current is , W == = . Alternatively, the integral of current is the final charge q on the capacitor, which equals CVSRC. The total energy delivered by the battery is W = qVSRC = . Version 1: VSRC = 10 V, W = 10-2100 = 1 J Version 2: VSRC = 11 V, W = 10-2121 = 1.21 J Version 3: VSRC = 12 V, W = 10-2144 = 1.44 J Version 4: VSRC = 13 V, W = 10-2169 = 1.69 J Version 5: VSRC = 14 V, W = 10-2196 = 1.96 J 3% 10. The switch is opened at t = 0, with no energy initially stored in the circuit. Determine vS(t), t  0+, assuming ISRC = 1 A dc. Solution: , , vS(t) = vC(t) + vL(t) = 5ISRC. Version 1: ISRC = 1 A, vS(t) = 5ISRC = 5 V Version 2: ISRC = 1.5 A, vS(t) = 5ISRC = 7.5 V Version 3: ISRC = 2 A, vS(t) = 5ISRC = 10 V Version 4: ISRC = 2.5 A, vS(t) = 5ISRC = 12.5 V Version 5: ISRC = 3 A, vS(t) = 5ISRC = 15 V 3% 11. The switch is closed at t = 0 with no energy initially stored in the circuit. Determine the largest value of di/dt, assuming VSRC = 2 V. p5/8 + – 10  10 mF VSRC t = 0 0.1 F t = 0 5  ISRC 5  2.5 H + – vS 10  1 mH 1 mF + – i t = 0 VSRC Solution: The largest value of di/dt occurs at t = 0, when vL = Ldi/dt = VSRC, and di/dt = VSRC,/L = 103VSRC A/s = VSRC A/ms, where VSRC is in V. Version 1: VSRC = 2 V, dit/dt = VSRC = 2 A/ms Version 2: VSRC = 3 V, dit/dt = VSRC = 3 A/ms Version 3: VSRC = 4 V, dit/dt = VSRC = 4 A/ms Version 4: VSRC = 5 V, dit/dt = VSRC = 5 A/ms Version 5: VSRC = 6 V, dit/dt = VSRC = 6 A/ms 3% 12. When the switch is opened at t = 0, with no energy initially stored in the circuit, it is found that v(t) = 2t V, t  0+ t is in s. Determine ISRC. Solution: At t = 0+, ISRC flows in C; iC(t) = Cdv/dt = Cd(Kt)/dt = CK = CK at t = 0+. It follows that ISRC = CK = 0.5K. Version 1: K = 2 V/s, ISRC = 1 A Version 2: K = 3 V/s, ISRC = 1.5 A Version 3: K = 4 V/s, ISRC = 2 A Version 4: K = 5 V/s, ISRC = 2.5 A Version 5: K = 6 V/s, ISRC = 3 A 13% 13. Determine the average power dissipated in the load, assuming vSRC(t) = 10cos(t + 45) V and R = 2 . Solution: VSRC = 1045 V. Let V be the voltage across the primary winding, I be the load current, and IS be the source current, as shown. It follows that V = ZLI, and V = VSRC – ZSIS (1), where ZL = (R + j10) , and ZL = (4R – j10) . From the mmf equation: NIS + NI – 2NI = 0, which gives IS = I. Substituting for V and IS in (1) and solving for I: A. P = W. Note that in this case, because the same current flows in the two secondary windings, the transformer can be considered as a 1:1 ideal transformer without voltage inversion; that is, the voltage across the primary winding is the load voltage, and the current in the primary winding is the load current. It follows that , as before. p6/8 + – v ISRC 0.5 F 1 S 0.5 H t = 0 R j10  – + 4R -j10  VSRC IS I V + _ V + _ 2V + _ + V _ N turns 2N turns N turns R j10  Load – + 4R -j10  VSRC Version 1: R = 2 ; P = 1 W Version 2: R = 4 ; P = 0.5 W Version 3: R = 8 ; P = 0.25 W Version 4: R = 10 ; P = 0.2 W Version 5: R = 20 ; P = 0.1 W 13% 14. Derive TEC looking into terminals ‘ab’, where VTH = Vab and ISRC = 0.10 A. Solution: Replacing the two coupled coils by the T- equivalent circuit, the circuit becomes as shown. 100I – 200I = 0, so that I = 0. It follows that Va = j30ISRC, and V1 = -j50ISRC. Hence, Vb = -2V1 = j100ISRC. This gives VTh = j30ISRC – j100ISRC = -j70ISRC. To determine ZTh, a test current source IT is applied, with the source ISRC replaced by an open circuit. The circuit becomes as shown. 100I – 200(I – IT) = 0, so that I = 2IT. From KVL around the outer loop, V1 = j30(I – IT) + j20I = j70IT. From KVL around the upper mesh, Vab = (-j10)IT + j20I + V1. Substituting for I and V1 in terms of IT, Vab = j30IT + j70IT = j100IT. this gives Vab/IT = ZTh = j100 . Version 1: ISRC = 0.10 A; VTh = -j70ISRC = 7-90 V Version 2: ISRC = 0.20 A; VTh = -j70ISRC = 14-90 V Version 3: ISRC = 0.30 A; VTh = -j70ISRC = 21-90 V Version 4: ISRC = 0.40 A; VTh = -j70ISRC = 28-90 V Version 5: ISRC = 0.50 A; VTh = -j70ISRC = 35-90 V 13% 15. Switch S1 is closed at t = 0, with the capacitor initially uncharged. Switch S2 is p7/8  ISRC a b 100 turns 200 turns j20  j30  -j10  V1 + _ _ 2V1 + + ISRC _ V1 I = 0 a b 100 turns 200 turns j20  j30  -j10  V1 + _ _ 2V1 + + _ V1 I  I – IT IT + – 2m A 20 k 5 k 15 V 0.5 F vC + _ t = 0 t = 30 ms 20 k S1 S2 S3 t = 50 ms  ISRC a b 100 turns 200 turns j10  j20  j10  closed at t = 30 ms, and switch S3 is closed at t = 50 ms. Determine vC(t) for all t  0+. Solution: vC(0+) = 0, VCF1 = 210-320103 = 40 V; 1 = 201030.510-6  10 ms. Hence, vC(t) = V, 0  t  30 ms. At t = 30 ms, vC(30ms) = = 38.0 V; the circuit seen by the capacitor for 30 ms  t  50 ms is as shown. The 15 V source in series with 5 k is transformed to a 3 mA current source in parallel with 5 k. The two current sources are combined into a 5 mA source and the parallel resistance is (205)/25 = 4 k; hence VCF2 = 20 V. 2 = 40.5 = 2 ms. Closing S3 at t = 50 ms has no effect on vC. It follows that vC(t) = V, or vC(t) = V, t  30 ms. Closing S3 places the 20 k resistor across an ideal voltage source, which does not affect vC. 13% 16. The switch is opened at t = 0 after being closed for a long time: (a) determine R for critical damping (4 grades); using this value of R, and assuming VSRC = 6 V, determine: (b) vC(t), t  0+, t in ms; (6 grades); (c) the energy in J stored in the capacitor and inductor as functions of time, t  0+, t in ms (3 grades). Solution: (a) rad/s. , R = 0.11000 = 100 . (b) vC(t) = ; at t = 0+, vC(0+) = VSRC = A; ; iL(0+) = , , which gives B = 0. Hence, , t  0+, t is in s, or, , t  0+, t is in ms. The response is that of a first-order circuit. (c) Energy stored in C: J, t  0+, t is in ms, , , t is in s; energy stored in L: , , t  0+, t is in ms. The energy stored in the capacitor and inductor is the same. The inductor current is in a direction to discharge the capacitor. Hence, the two stored energies decay together, like a single stored energy. This is the fundamental reason behind the first-order response. Version 1: VSRC = 2 V, = 20 Version 2: VSRC = 3 V, = 45 Version 3: VSRC = 4 V, = 80 Version 4: VSRC = 5 V, = 125 Version 5: VSRC = 6 V, = 180 p8/8 + – t = 0 10 F 0.1 H R R VSRC + – vC + – 2m A 20 k 5 k 15 V vCF2 + _ + – t 0+ 10 F 0.1 H R R VSRC + – vC iL iC