EECE 210 Final, Fall 2009

EECE 210 Final Exam fall 2009/2010 – Sections 1, 3 and 4 Closed Book – Three Hours 3% 1. Determine the power dissipated in the circuit, assuming I = 1 A. Solution: The 1  Y is paralleled with a 3  , so that it effectively becomes a 0.5  Y, and the circuit reduces to that shown. The resistance seen by the current source is 1||1 + 2.5 = 3 , so that the power dissipated in the circuit is P = 3I2 W. Version 1: I = 1 A, P = 3 W Version 2: I = 2 A, P = 12 W Version 3: I = 3 A, P = 27 W Version 4: I = 4 A, P = 48 W Version 5: I = 5 A, P = 75 W. 3% 2. Determine the power delivered by the 3 V source, assuming  = 2 V/A. Solution: The upper node is at 5 V with respect to the lowest node, the middle node is at 3 V. hence, Ix = 0.5 A and the current in the 6  resistor is also 0.5 A. The current supplied by the 3 V source is (3 – 0.5)/2 and the power delivered by the source is P = 1.5(3 – 0.5) = 4.5 – 0.75 W. Version 1:  = 2, P = 4.5 – 0.752 = 3 W Version 2:  = 3, P = 4.5 – 0.753 = 2.25 W Version 3:  = 4, P = 4.5 – 0.754 = 1.5 W Version 4:  = 5, P = 4.5 – 0.755 = 0.75 W Version 5:  = 6, P = 4.5 – 0.756 = 0 3% I 1  1  1  1  1  3  3  2  2  0.5  I 1  0.5  0.5  0.5  2  0.5  – + – + – + 10  4  6  2  5 V 3 V 10 A Ix Ix – + 40 V 5  j10  2:1 5  -j5  1:a 1/8 3. Determine the impedance seen by the source, assuming a = 2. Solution: Reflection of the (5 – j5)  through the RH transformer gives (20 – j20) . The impedance on the secondary side of the LH transformer is (25 – j10) . Reflected to the primary side, this becomes (25 – j10)/a2 . Version 1: a = 2, Z = 6.25 – j2.5  Version 2: a = 3, Z = 2.78 – j1.11  Version 3: a = 4, Z = 1.56 – j0.63  Version 4: a = 5, Z = 1 – j0.4  Version 5: a = 6, Z = 0.69 – j0.28  3% 4. If vsrc = 10cos(1,000t) V, determine the energy stored in the circuit in the sinusoidal steady state at t = 0, assuming C = 1 F. Solution: At t = 0, the voltage across C is 10 V and the current through the inductors is zero, being proportional to the integral of vsrc. The energy stored is W = 50C. Version 1: C = 1 F, W = 50 J Version 2: C = 2 F, W = 100 J Version 3: C = 3 F, W = 150 J Version 4: C = 4 F, W = 200 J Version 5: C = 5 F, W = 250 J. 3% 5. Determine Rx given that I = 0 and R = 2 . Solution: Since I = 0, the voltage across Rx is 10 V, and the same current flows through R and Rx. It follows that , or . Version 1: R = 2 , Rx = 1  Version 2: R = 3 , Rx = 1.5  Version 3: R = 4 , Rx = 2  Version 4: R = 5 , Rx = 2.5  Version 5: R = 6 , Rx = 3  1H 1H k = 0.5 – + vsrc C – + 100 V 8  6  -j5  – + 300 V -j10  R Rx I 2/8 3% 6. Determine the power absorbed or delivered by the dependent source assuming R = 1 . Solution: The current in the 2  resistor is 2Ix flowing downwards. From KVL in the mesh on the left, 10 = 4Ix + 2, or Ix = 2 A. The voltage rise Vx across the dependent source is given by: Vx – RIx = 5, or Vx = 2R + 5; The power P delivered by the source is P = 2(2R + 5). Version 1: R = 1 , P = 2(21 + 5) = 14 W Version 2: R = 2 , P = 2(22 + 5) = 18 W Version 3: R = 3 , P = 2(23 + 5) = 22 W Version 4: R = 4 , P = 2(24 + 5) = 26 W Version 5: R = 5 , P = 2(25 + 5) = 30 W. 3% 7. Determine the maximum power that can be delivered to RL, assuming R = 0.5 . Solution: The primary voltage of the upper transformer is always 1 V. On open circuit, the source current is zero, the primary voltage is 5 – 1 = 4 V, and VTh = 8 V. On short circuit, the primary voltage of the lower transformer is zero, the source current is (5 – 1)/R and the short circuit current is 2/R. This gives, RTh = 4R. The maximum power delivered is (8)2/(44R) = 4/R. Version 1: R = 0.5 , P = 4/0.5 = 8 W Version 2: R = 1 , P = 4/1 = 4 W Version 3: R = 2 , P = 4/2 = 2 W Version 4: R = 2.5 , P = 4/0.8 = 1.6 W Version 5: R = 4 , P = 4/4 = 1 W. 3% 8. Given that the load L consumes 1200 W at 0.8 p.f. lagging and the magnitude of the voltage – + 10 V R 10  2  – + 2 V – + Ix 5 V Ix -j30  L RLine Supply voltage – + 50 V R – + 20 V 1:2 1:2 RL 3/8 across L is 300 V rms. Determine the power dissipated in the resistance Rline, if Rline = 0.5 . Solution: The reactive power absorbed by the load is VAR. The reactive power absorbed by the capacitor is VAR. The total complex power is 1200 + j(900 – 3000) = 1200 – j2100 VA. The magnitude of the line current A. The power dissipated in Rline is 65Rline. Version 1: R = 0.5 , P = 650.5 = 32.5 W Version 2: R = 0.6 , P = 650.6 = 39 W Version 3: R = 0.7 , P = 650.7 = 45.5 W Version 4: R = 0.8 , P = 650.8 = 52 W Version 5: R = 0.9 , P = 650.9 = 58.5 W. 3% 9. Two identical coils, each having an inductance of 10 mH, are connected in series. When the connections to one of the coils are reversed, the total inductance is multiplied by a factor a. Determine the coupling coefficient of the coils. Solution: (10 + 10 + 2M) = a(10 + 10 – 2M); 2M(a + 1) = 20(a – 1); ; Version 1: a = 1.5, 0.2 Version 2: a = 1.6, 0.23 Version 3: a = 1.7, 0.26 Version 4: a = 1.8, 0.29 Version 5: a = 1.9, 0.31. 3% 10. Determine Ix, assuming R = 4 . Solution: The voltage across all windings is zero. Hence, I1 = A, and I2 = A. Setting the net mmf to zero, 400I1 – 100I2 + 200I3 = 0, or I3 = 0, which gives I3 = ; IX = I2 – I3 = . Version 1: R = 4 ; Ix = 6 A Version 2: R = 5 ; Ix = 5 A Version 3: R = 8 ; Ix = 3.5 A Version 4: R = 10 ; Ix = 3 A Version 5: R = 20 ; Ix = 2 A. 3% – + 100 V R 2  – + 400 turns 100 turns 200 turns 40 V Ix – + 100 V R 2  – + 400 turns 100 turns 200 turns 40 V Ix I1 I2 I3 – + L 8  10  1 F 50 V 4/8 11. Determine the frequency at which maximum power is dissipated in the 10  resistor, assuming L = 1 H. Solution: . Maximum power is dissipated in the 10  resistor when XL = -XC, which gives L = , or rad/s. Version 1: L = 1 H; rad/s Version 2: L = 2 H; = 0.71rad/s Version 3: L = 3 H; = 0.58 rad/s Version 4: L = 4 H; rad/s Version 5: L = 5 H; = 0.45 rad/s 3% 12. For n = 1, 2, 3,…, the function shown has: A. an and bn nonzero for all n B. an and bn are zero for even n C. an and bn are zero for odd n D. an = 0 for all n E. bn = 0 for all n Solution: When the dc value is removed, the ac component has half-wave symmetry but is neither even nor odd. Hence, an and bn are zero for even n. 3% 13. Determine the total power dissipated if vI is a full-wave rectified waveform given by: vI = 6|sin(500t)| V. Solution: ohms; ; ; ; ; . Version 1: Vm = 6 V, W Version 2: Vm = 7 V, 2.91 W Version 3: Vm = 8 V, 3.81 W Version 4: Vm = 9 V, 4.82 W Version 5: Vm = 10 V, 5.95 W. 3% 14. A period of a periodic function f(t) is given by: K(4 + 2sint), 0 < t < 2. Determine the rms value of f(t), if K = 0.5. Solution: The square of f(t) is 100 F 5  5  – + vI f(t) t T -T/2 T/2 A A f(t) t  4 2 2 5/8 . The area under the square is ; the mean square is and the rms value is . Version 1: K = 0.5, rms = = 2.12 Version 2: K = 0.6, rms = = 2.55 Version 3: K = 0.7, rms = = 2.97 Version 4: K = 0.8, rms = = 3.39 Version 5: K = 0.9, rms = = 3.82. 3% 15. Determine the total power dissipated in R if R = 1 . Solution: With either of the 10cos100t V acting alone, iR1 = . The current due to both 10cos100t V sources, with the 10cos200t V source set to zero, is and the power dissipated in R is . With the 10cos200t V source acting alone, iR3 = , and the power dissipated in R is . The total power dissipated in R is . Version 1: R = 1 ; P = W Version 2: R = 1.5 ; P = 24 W Version 3: R = 2 ; P = 22.22 W Version 4: R = 2.5 ; P = 20.41 W Version 5: R = 3 ; P = 18.75 W. 11% 16. Determine VO. Solution: The 2Vx source is replaced by a 10 A source. The current in the 2  resistor is Ix. The current in the dependent source is 5 – 2Ix, so that the current in the 1  resistor is 15 – Ix. From KVL around the mesh abcd, 2Ix + 15 – Ix = 4Ix, which gives Ix = 5 A. It follows that VO = 15 – Ix = 10 V. – + 5 A 2Ix 2Vx 2  4  1  Ix – + Vx – + VO 1  2  2  – + R – + 10cos100t V – + 10cos100t V 10cos200t V – + 5 A 10 A 2  4  1  – + VO Ix 5 - 2Ix 1  a Ix 2Ix 2Ix 15 - Ix b d c 6/8 11% 17. Determine iO, given that VSRC1 is 15 V dc and VSRC2 = 10cos(3,000t) V. Solution: With VSRC1 applied and VSRC2 set to zero, the circuit becomes as shown. 15 = 3Vx, so that Vx = 5 V and A. With VSRC2 applied and VSRC1 set to zero, the circuit becomes as shown. It follows that: -2Vx = Vx + , or , which gives Vx = 0. Hence, IO2 A. Thus, iO = -1 + 2cos(3,000t) A. 11% 18. Determine k so that the input resistance is purely resistive. Solution: Disregarding the 10  resistance and replacing the linear transformer by its T-equivalent circuit, the circuit becomes as shown. The input reactance is , or , which gives . Hence, 0.71. 11% – + 2Vx 2  5  – + Vx 5  2  15 V – + IO1 – + 2Vx 2  5  – + Vx 100 V 5  – + 2  jXL jXC IO2 k j40  j10  -j20  10  Zin -j20  j(40 + XM)  j(10 + XM)  -jXM  + 100 V 5  j2  5  Z – + 2Vx 2  5  – + Vx VSRC2 5  – + 2  10 mH 50 F VSRC1 – + iO 7/8 19. Determine Z so that maximum power is transferred to it and calculate this power given that the source voltage is 10 V peak value. Solution: We will determine TEC as seen by Z. On open circuit, the currents are as shown. From KVL: 100 – 5I/2 + 5I/2 = VTh. In This particular problem, the voltages across the 5  resistors cancel out. Hence, VTh = 100 V peak value When Z is replaced by a short circuit, the currents are as shown. From KVL: 100 – 5(Isc + I/2) – 5(Isc – I/2) = 0. Again, the terms involving I cancel out. Hence, Isc = 10 A, and ZTh = . It follows that for maximum power transfer, Z = 10 . The power dissipated in the load is W. 11% 20. Derive the trigonometric Fourier expansion of the given periodic function f(t). Solution: Since f(t) is odd, a0 = 0 = an; T = 2, 0 = 2/T = ; f(t) = t + 1; = = = = f(t) =. 1 -1 2 -2 1 2 -1 -2 t f(t) – + 100 V 5  j2  1:2 5  – + VTh I I/2 I/2 I/2 – + 100 V 5  j2  1:2 5  I I/2 ISC – I/2 Isc Isc + I/2 8/8