EECE 210 Final Exam-solution, Fall 2016

EECE 210 Electric Circuits Final Exam – Dec 14, 2016 12% 1. Using superposition , determine: (a) VO (8 grades) and (b) the power delivered or absorbed by the independent voltage source (4 grades). Solution: The dependent source is replaced by an independent source VX = 2VO. With the 2 V source applied alone, VO1 = 0. With the 2 A current source applied alone, the inductor is short circuited, and ; with VX applied alone, it follows from voltage division that . Substituting for VX, the superposition equation is: VO = (1 – j) + VO(1 + j), which gives VO = (1 + j) V. The voltage across the dependent source is (2 + j2) V; the voltage across the inductor is 2 + j2 – 2 = j2 V, and the current flowing through the inductor towards the 2 V source is j2/j = 2 A. The current through the 2 V source is zero, so that this source neither delivers nor absorbs power. 12% 2. Determine the open-circuit voltage Vab and the short-circuit current Iab , as phasors in polar coordinates, assuming vSRC = 3cos103t V. Solution: Replacing the coupled coils by the T- equivalent circuit, the circuit becomes in the frequency domain as shown. It follows from voltage division that: . On short circuit, the -j3  and j3  branches act as an open circuit, so that the current toward these branches is zero and the voltage across them is VSRC. It follows that A. Version 1: VSRC = 30; Vab = 4.5-90 V, Iab = 1-90 A. Version 2: VSRC = 60; Vab = 9-90 V, Iab = 2-90 A. Version 3: VSRC = 90; Vab = 13.5-90 V, Iab = 3-90 A. Version 4: VSRC = 120; Vab = 18-90 V, Iab = 4-90 A. Version 5: VSRC = 150; Vab = 22.5-90 V, Iab = 5-90 A. 1/5 1 mH 1 mH 2 mH 1 mH 2  0.5 mF vSRC – + a b -j  j3  2  -j2  VSRC – + a b j3  Iab I = 0  – + – + + – VO 1  -j  j  2 A 2 V 2VO -j  j3  2  -j2  VSRC – + a b j3  12% 3. (a) Transform the 100 A source and its source impedance to its equivalent voltage source, and (b) determine VO using the mesh- current method, assuming ISRC = 10 A. Solution: Mesh 1: I1 = ISRC; Mesh 3: -j2I2 + (8 + j2)I3 = 0 For the dependent source, IX = I3 = I2 – I1 = I2 – ISRC, so that I2 = I3 + ISRC. Substituting in the equation for mesh 3: -j2(I3 + ISRC) + (8 + j2)I3 = 0, or, -j2ISRC + 8I3 = 0, or, I3 = j0.25ISRC, so that VO = 8I3 = j2ISRC V. Version 1: ISRC = 10 A; VO = 290 V. Version 2: ISRC = 20 A; VO = 490 V. Version 3: ISRC = 30 A; VO = 690 V. Version 4: ISRC = 40 A; VO = 890 V. Version 5: ISRC = 50 A; VO = 1090 V. 13% 4. Determine VS, assuming ISRC = 10 A. Solution: From symmetry the –j20 reactance does not carry current. It is redundant and can be removed. Replacing the coupled coils by the T- equivalent circuit, the circuit becomes as shown. it follows that the impedance seen by the source Is –j17.5 + j5 + j45/2 = j10 , so that VS = j10ISRC. Alternatively, transforming the delta to a star, and replacing the coupled coils by the T-equivalent circuit, the circuit becomes as shown. It is seen that the impedance seen by 2/5 ISRC 4  100 A - j 2  8  IX IX + – VO j 2    ISRC 4  -j20 V - j 2  8  IX IX + – VO j 2   I1 I2 I3 –  +  ISRC -j17.5  j15  j15  + – VS j5  j30  j30  the source is -j17.5 + j5 + j22.5 = j10 . It follows that VS = j10ISRC. Version 1: ISRC = 10; VS = 1090 V. Version 2: ISRC = 1.50; VS = 1590 V. Version 3: ISRC = 20; VS = 2090 V. Version 4: ISRC = 2.50; VS = 2590 V. Version 5: ISRC = 30; VS = 3090 V. 13% 5. Determine IL as a phasor in polar coordinates, assuming vSRC(t) = cos(t + 45) V. Solution: L = 22.5 , 1/C = 10 ; the circuit in the frequency domain is as shown. V1 = VSRC – j45IL; V2 = VSRC + j30IL; 2V1 + 3V2 = 10IL; substituting for V1 and V2, 2VSRC – j90IL + 3VSRC + j90IL = 10IL; IL = 0.5VSRC A Version 1: VSRC = 145; IL = 0.545 A. Version 2: VSRC = 245; IL = 145 A. Version 3: VSRC = 345; IL = 345 A. Version 4: VSRC = 445; IL = 245 A. Version 5: VSRC = 545; IL = 2.545 A. 12% 6. The switch is closed at t = 0, with no stored energy in the circuit. Determine iS(t), t  0+, assuming VSCR = 10 V. Solution: The current flowing downwards in the capacitive branch is A. The current flowing downwards in the inductive branch is A, where C = 10×0.001= 0.01 s and L = 0.1/10 = 0.01 s. It follows that iS(t) = VSRC/10 A. 3/5 + – 10  10  0.1 H 1 mF VSRC iS 1:2 1:3 – + 10  22.5 H 0.1 F vSRC iL 1:2 1:3 – + 10  VSRC IL V1 _ + 2V1 _ + V2 _ + 3V2 _ + 3IL 2IL -j10  22.5  Version 1: VSRC = 10, iS(t) = 10/10 = 1 A. Version 2: VSRC = 15, iS(t) = 15/10 = 1.5 A. Version 3: VSRC = 20, iS(t) = 20/10 = 2 A. Version 4: VSRC = 25, iS(t) = 25/10 = 2.5 A. Version 5: VSRC = 30, iS(t) = 30/10 = 3 A. 13% 7. Both switches are moved at t = 0 after being in their initial positions for a long time. Determine vC(t) and iL(t) for t  0+. Solution: At t = 0-, it follows from current division that iL(0-) = 10 A and the current through the 4  resistor is 20 A, so that vC(0-) = 80 V. These do not change at t = 0+. As t  , the capacitor acts as an open circuit and the inductor as a short circuit, so that iL() = 0. The 5 A current flows through the 8  resistor, making vC() = 40 V. When the 5 A source is set to zero, the circuit is a series circuit;  = R/2L = 8/20 = 0.4 rad/s; rad/s; the responses are underdamped, and rad/s. V; At t = 0+, 80 = A + 40, so that A = 40 V; at t = 0+, iL = -CdvC/dt = -C(-A + dB) = 10; dB = -25 + 0.440 = -9, B = -9/0.3 = -30. Hence, . iL(t) = -CdvC/dt A. 13% 8. Switch S1 is moved from position ‘a’ to position ‘b’ at t = 0 after being in position ‘a’ for a long time. Switch S2 is closed at t = 0.5545 ms. Determine iL(t) and vC(t) for t  0+. Solution: At t = 0, vX = 0 and vC(0+) = 5×4 = 20 V; 0  t  0.5545 ms,  = 100×8×10-6 = 8×10-4  0.8 ms, and vC(t) =. At t = 0.5545 ms, vC(t) = 10 V. Let t = t – 0.5545 ms; For t > 0, the series is a parallel GCL circuit having 0 = rad/s; p = rad/s. The responses are overdamped, with s1 = -1250 + -1250 + 4/5 – + + – vC + – vX 2vX 8 F 100  100  0.125 H a b S1 4  5 A 10  iL + – 10 V S2 8  10 H t = 0 0.4 F 4  30 A iL + – vC t = 0 a b 5 A 750 = -500 rad/s and s2 = -1250 – -1250 – 750 = -2000 rad/s. IL0 = 0 and ILF = -10/100 = -0.1 A. Hence, A; at t = 0, A + B = 0.1; At t = 0+, vC(0+) = 10 = 10 + LdiL/dt. Hence, LdiL/dt = 0 at t = 0+, or -500A – 2000B = 0. Substituting A = -4B gives B = -0.1/3 A and A = 0.4/3 A. Hence, mA, where t is in ms. = + 10 V. 5/5   2 2 (1000) (1250)