EECE 210 Exam 3 with Solution, Fall 2014

EECE 210 Quiz 3, November 21, 2014 4% (Primal Question) A. Determine vO(t), assuming vSRC(t) = cos10t V. Solution: jL = j , ; V. Version 1: vSRC(t) = cos10t V; V Version 2: vSRC(t) = 2cos10t V; V Version 3: vSRC(t) = 3cos10t V; V Version 4: vSRC(t) = 4cos10t V; V Version 5: vSRC(t) = 5cos10t V; V. 4% (Primal Question) B. Given v1(t) = 12sin(t + 60) V and v2(t) = -6cos(t + 30) V. Express v1(t) as a cosine function, convert v1(t) and v2(t) to phasors, and draw them on a phasor diagram. Solution: cos(t – 90) = sint; hence, v1(t) = 12cos(t – 30) V; V1 = 12-30 V, V2 = -630 = 6210 V. Version 1: v1(t) = 12cos(t – 30) V and v2(t) = 6cos(t + 210) V Version 2: v1(t) = 6cos(t – 30) V and v2(t) = 6cos(t + 210) V Version 3: v1(t) = 6cos(t – 30) V and v2(t) = 12cos(t + 210) V Version 4: v1(t) = 12cos(t – 60) V and v2(t) = 6cos(t + 240) V Version 5: v1(t) = 6cos(t – 60) V and v2(t) = 12cos(t + 240) V. 4% (Primal Question) C. Determine VO, given VSRC = 10 V, L1 = 10 , L2 = 40 , and k = 0.5. Express VO in rectangular coordinates (of the form a + jb). Solution: M = 0.5 . The current 1 – + L1 L2 k 10  VSRC0 VO + _ – + 0.1 H 1  vO + _ vSRC Re Im V1 V2 entering the dotted terminal of L1 is: ; hence, 0.5(1 + j) V. Version 1: VSRC = 10 V, VO = 0.5(1 + j)VSRC = 0.5(1 + j) V Version 2: VSRC = 20 V, VO = 0.5(1 + j)VSRC = (1 + j) V Version 3: VSRC = 30 V, VO = 0.5(1 + j)VSRC = 1.5(1 + j) V Version 4: VSRC = 40 V, VO = 0.5(1 + j)VSRC = 2(1 + j) V Version 5: VSRC = 50 V, VO = 0.5(1 + j)VSRC = 2.5(1 + j) V. 4% (Primal Question) D. Determine iO(t), given that vSRC(t) = 5sint V. Solution: vO = -2vSRC and iO = -vO/10. Hence, iO = 0.2vSRC. Version 1: vSRC(t) = 5sint A, iO(t) = 0.2vSRC(t) = sint A Version 2: vSRC(t) = 10sint A, iO(t) = 2vSRC(t) = 2sint A Version 3: vSRC(t) = 15sint A, iO(t) = 0.2vSRC(t) = 3sint A Version 4: vSRC(t) = 20sint A, iO(t) = 0.2vSRC(t) = 4sint A Version 5: vSRC(t) = 25sint A, iO(t) = 0.2vSRC(t) = 5sint A. 4% 1. Determine the equivalent impedance between terminals ‘ab’ assuming all capacitances are 1/3 F, all inductances are 1 H, and  = 1 rad/s (Note: when impedances are connected in delta or star, Z = 3ZY for equivalence). A. -j4  B. -j2  C. -j  D. -j5  E. -3  Solution: The impedance of each capacitor is -j/(1C) = -j/C . The impedance of each of the equivalent Y-connected capacitors is -j/(3 C )  = -jL , since the numerical value of L is 1/(3C). The impedance of each inductor is +j(1L) = jL . The impedance -jL + jL = 0 in the two parallel branches. Hence, Zeq = -jL . Version 1: C = 1/3 F, L = 1 H, Zeq = -jL = -j . Version 2: C = 1/6 F, L = 2 H, Zeq = -jL = -j2 . Version 3: C = 1/9 F, L = 3 H, Zeq = -jL = -j3 . 2 a b b a     3 jL C j +jL  vO + _ 1:2 10  vSRC iO – + – + vSRC + vO _ 1 H 1 H R R Version 4: C = 1/12 F, L = 4 H, Zeq = -jL = -j4 . Version 5: C = 1/15 F, L = 5 H, Zeq = -jL = -j5 . . 4% 2. If the power dissipated in the 5  resistor is 1 W, determine the power dissipated in the dual of the 5  resistor in the circuit that is the dual of the given circuit. A. 1 W B. 4 W C. 3 W D. 2 W E. 5 W Solution: By definition of duality, voltage and current are interchanged in each element in a given circuit and its dual in the dual circuit. The power therefore remains the same. Version 1: P = 1 W Version 2: P = 2 W Version 3: P = 3 W Version 4: P = 4 W Version 5: P = 5 W. 4% 3. Determine , assuming vSRC = 2cost and vO = 2sint and R = 2 . A. 4 rad/s B. 5 rad/s C. 6 rad/s D. 2 rad/s E. 3 rad/s Solution: ; -90 = -2tan-1(L/R); tan(L/R) = 1;  = R/L = R. Version 1: R = 2 ,  = 2 rad/s Version 2: R = 3 ,  = 3 rad/s Version 3: R = 4 ,  = 4 rad/s Version 4: R = 5 ,  = 5 rad/s Version 5: R = 6 ,  = 6 rad/s. 3 + – VSRC 1 H 5  2 F 10  – + j5  -j3  j5 V j10  j  j2  10:1 -j10  5  ISRC a b j0.2  4% 4. Determine ZTh seen between terminals ‘ab’, assuming Zc = -j . A. -j3  B. -j4  C. -j5  D. -j2  E. -j  Solution: when a test source is applied between terminals ‘ab’ and independent sources are set to zero, the circuit reduces to that shown. IX = 0, so that the independent source can be replaced with an open circuit. It follows that ZTh = Zc. Version 1: Zc = -j , ZTh = -j  Version 2: Zc = -j2 , ZTh = -j2  Version 3: Zc = -j3 , ZTh = -j3  Version 4: Zc = -j4 , ZTh = -j4  Version 5: Zc = -j5 , ZTh = -j5 . 4% 5. Determine Zeq, assuming Z = 3 + j4  and Xc = 0.5 . A. j2.25  B. j2.75  C. j2.5  D. j1.75  E. j2  Solution: Replacing the coupled coils by the T-equivalent circuit shows a balanced bridge, so that Z could be replaced by an open circuit or a short circuit. Replacing it by a short circuit, gives: Zeq = j2 + j1 – jXc/2 = j3 – jXc/2 . Version 1: Xc = 0.5 , Zeq = j3 – jXc/2 = j2.75  Version 2: Xc = 1 , Zeq = j3 – jXc/2 = j2.5  Version 3: Xc = 1.5 , Zeq = j3 – jXc/2 = j2.25  Version 4: Xc = 2 , Zeq = j3 – jXc/2 = j2  Version 5: Xc = 2.5 , Zeq = j3 – jXc/2 = j1.75 . 4% 4 j2  j4  j4  Z -jXc -jXc Zeq Zeq a b j2  j2  j2  -jXc -jXc -j3  2  IX VSRC ISRC j5  IX j2  Zc j2  190 A a b -j3  2  IX j7  IX Zc a b 6. Determine VTh = Vab, assuming ISRC = (1 + j) A. A. 6(-1 + j) V B. 3(-1 + j) V C. 4(-1 + j) V D. 2(-1 + j) V E. 5(-1 + j) V Solution: Because of the series current source, the circuit reduces to that shown, after reflection to the output side. The parallel impedance is: ; VTh = -(-j0.2*10ISRC) = j2ISRC V. Version 1: ISRC = (1 + j) A, VTh = j2ISRC = 2(-1 + j) V Version 2: ISRC = 1.5(1 + j) A, VTh = j2ISRC = 3(-1 + j) V Version 3: ISRC = 2(1 + j) A, VTh = j2ISRC = 4(-1 + j) V Version 4: ISRC = 2.5(1 + j) A, VTh = j2ISRC = 5(-1 + j) V Version 5: ISRC = 3(1 + j) A, VTh = j2ISRC = 6(-1 + j) V. 20% 7. Determine iC(t), assuming vSRC(t) = sin(1000t + 30) V. Solution: jL = j10000.1 = j100 ; -j/C = -j/(100010-5) = -j100 ; it follows that the parallel impedance of L and C is infinite. The circuit reduces to that shown in the frequency domain. From KVL, VSRC = 5IX + 5(8 + 1)Ix = 50Ix; Ix = VSRC/50 mA; VC = 1(5IX) = VSRC/10 V; IC = jCVC = jVC/100 A = jVSRC mA = j130 = 1120 mA. Hence, iC(t) = sin(1000t + 120) mA. Version 1: vSRC(t) = sin(1000t + 30) V, iC(t) = sin(1000t + 120) mA Version 2: vSRC(t) = 2sin(1000t + 30) V, iC(t) = 2 sin(1000t + 120) mA Version 3: vSRC(t) = 3sin(1000t + 30) V, iC(t) = 3 sin(1000t + 120) mA Version 4: vSRC(t) = 4sin(1000t + 30) V, iC(t) = 4 sin(1000t + 120) mA Version 3: vSRC(t) = 5sin(1000t + 30) V, iC(t) = 5 sin(1000t + 120) mA. 20% 5 a b j0.2  -j0.1  10ISRC 1 k 10 F 0.1 H – + 8 k 5 k 4iX vSRC iC iX 1 k – + 8 k 5 k 4IX VSRC IX VC + _ 5IX V1 + _ – + 1:2 Io 0.5Io 10  30  VSRC 0.5Io Io 2V1 + _ 1.5Io 8. (a) Derive the mesh-current equations (15 grades); (b) determine Vab (5 grades). Solution: Mesh 1: j20I1 + j10I2 – j10I4 = VY, where VY is a voltage rise across the dependent source in the direction of I1. Mesh 2: j10I1 + (j10 – j10)I2 + – (-j10I3) = -VY. Adding these: j30I1 + j10I2 + j10I3 – j10I4 = 0; Substituting I4 = -j and I3 = 1 and dividing by j10: 3I1 + I2 = -1 – j. Also, VX = I2 – I1, where Vx = 3(1 + j). Substituting for I2: -4I1 = (1 + j) + 3(1 + j) = 4(1 + j), or -I1 = (1 + j). Vab = j10(I4 – I1) = j10( -j + 1 + j) = j10 V. 20% 9. Determine the power dissipated in the 30  resistor given that vSRC = 6sin(t + 30) V. Solution: The circuit is represented in the frequency domain as shown, with the voltages and currents assigned as indicated. From Ohm’s law on the secondary side, 2V1 = 30Io, or V1 = 15Io. From KVL on the primary side, VSRC = 15Io + V1. Substituting for V1, VSRC = 30Io, or Io = VSRC/30. Power dissipated in 30  resistor is P = W. Version 1: vSRC = 6sin(t + 30) V, P = 60 2 VSRC = 36/60 = 0.6 W Version 2: vSRC = 12sin(t + 30) V, P = 60 2 VSRC = 2.4 60 12 12   W. Version 3: vSRC = 18sin(t + 30) V, P = 60 2 VSRC = 5.4 60 18 18   W. Version 4: vSRC = 24sin(t + 30) V, P = 60 2 VSRC = 9.6 60 24 24   W. 6 – + 1:2 iO 0.5iO 10  30  vSRC VX j A b 3  1 A j10  j10  j10  + _ VX a I4 I1 I2 I3 j10  -j10  Version 5: vSRC = 30sin(t + 30) V, P = 60 2 VSRC = 15 60 30 30   W. 7