EECE 210 Homework 10 with Solution, Fall 2013

Homework 10 P9.2.9 Determine TEC looking into terminals ‘ab’ in Figure P9.2.9, assuming vSRC = 2cos(103t - 60) V and k = 0.75. Ans. 6120 V, -j0.5 . Solution: The impedances are: -j/C = -j/(103210-3) = -j0.5 ; jL1 = j(103110-3) = j ; jL2 = j(103410-3) = j4 ; jM = j(1030.75210-3) = j1.5 . The circuit in the frequency domain with the linear transformer replaced by its T- equivalent circuit is as shown. The series impedances are (jL1 + jM) = j2.5  and (jL2 + jM) = j5.5 , the shunt impedance being -jM = -j1.5 . VTh . If the source is replaced by a short circuit, the impedance looking into terminals ‘ab’ is = -j0.5 . P9.2.10 Determine in Figure P9.2.10. Ans. j20 . Solution: The impedance encountered by the current I is jL1 + jL2 – j 2M – j /C = j20 + j20 – j20 – j20 = 0. Hence, Zin = 0. P9.2.14 Determine the frequency at which the current in Figure P9.2.14 has the same magnitude when the connections of one coil are reversed. 1-10 vSRC k 1mH 4mH 2 mF a b Figure P9.2.9 – + – + Vm-60 -j0.5  a b j2.5  j5.5  -j1.5  j 20  j 10  j 20  j 10  - j 20  Figure P9.2.10 Zin j 20  j 10  j 20  j 10  - j 20  Zin I 30 mH 120 mH 6 F k = 0.75 + – v i Figure P9.2.14 Solution: M  0.75  45 mH. Since the current magnitude remains the same when one coil is reversed, the impedance reverses sign. Thus: j(30 + 120 + 90) –  -j(30 + 120 - 90) + , or 300  . 2  ;   krad/s. P9.2.16 Given vSRC = 6cost V and k = 0.9 in Figure P9.2.16. Determine X so that no power is dissipated in the circuit. Solution: For no power dissipation, the current in the 10  resistor should be zero. From the T- equivalent circuit, I2 = 0 if the shunt branch is a short circuit. This requires X = -M, where -0.920 = -18 . P9.2.19 in Figure P9.2.19 is initially charged to and is uncharged. The switch is closed at Calculate the total energy dissipated in the resistor. Solution: The energy that is eventually dissipated in the resistor is that initially stored in . This energy is 0.510–6(6)2 = 18 J. 2-10 k j10  10  VSRC Figure P9.2.16 j40  jX – + j(10 – M) 10  VSRC jM  jX – + j(40 – M) I2 15 mH 30 mH 18 mH 1 k 1 F 2 F C1 t = 0 C2 Figure P9.2.19 6  3  k = 0.5 9 H iO 4 H vSRC – + P9.2.20 Given vSRC = 200sin(103t) V in Figure P9.2.20. Determine . Solution: Considering I1 to be a mesh current and IO to be the current in the outer loop, KVL gives: (6 + j4000)I1 + (j3000 +6)Io = - j200, and (j3000 +6)I1 + (9 +j7000)Io = -j200. Solving, gives Io = -0.0105  -0.0105cos(1,000t) A. P9.2.23 Determine VO in Figure P9.2.23. Solution: Replacing the linear transformer by its T- equivalent circuit and writing the mesh- 3-10 6  – + 3  IO j 4,000  -j200 V j 9,000  -j 2,000  j 3,000  I1 j 5  j 8  j 10  20  10Ix Ix 2030 – + + – VO Figure P9.2.23 – + j 5  j 3  10Ix V + – + + V j 5  I3 I1 I2 current equations: j8I1 – j5I2 – j3I3  2030, -j5I1 + (20 + j10)I2 – j5I3  0, -j3I1 – j5I2 + j8I3  10(I1 – I2), or, -(10 + j3)I1 + (10 – j5)I2 + j8I3  0. Solving, I2  0.897 – j0.777 A; hence, VO = 20I2 = 17.94 – j15.53 V. P9.2.24 Given V in Figure P9.2.24. Determine . Solution: j500210–3 = j ; j500810–3 = j4 ; mH; j5003.210–3 = j1.6 ; 4-10 10 F 10  k = 0.8 5  8 mH 2 mH iO vSRC Figure P9.2.24 – + 10  5  – + IO j 2.4  -j 200  500 j 1.6  -j 0.6  I1 I2 -j/(5001010–6) = -j200 . Replacing the linear transformer by its T-equivalent circuit, the mesh current equations are: (5 + j4 – j200)I1 – (j1.6 – j200)I2 = 50; -(j1.6 – j200)I1 + (10 + j – j200)I2 = 0. This gives Io = I2 = 3.29 – j0.455 = 3.33-7.87  3.33cos(500t – 7.87) A. 5-10