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Bài giải phần giải mạch P13

Chapter 13, Solution 1. For coil 1, L1 – M12 + M13 = 6 – 4 + 2 = 4 For coil 2, L2 – M21 – M23 = 8 – 4 – 5 = – 1 For coil 3, L3 + M31 – M32 = 10 + 2 – 5 = 7 LT = 4 – 1 + 7 = 10H or LT = L1 + L2 + L3 – 2M12 – 2M23 + 2M12 LT = 6 + 8 + 10 = 10H Chapter 13, Solution 2. L = L1 + L2 + L3 + 2M12 – 2M23 2M31 = 10. | Chapter 13 Solution 1 For coil 1 Li - M12 Mn 6 - 4 2 4 For coil 2 L2 - M21 - M23 8 - 4 - 5 - 1 For coil 3 L3 M31 - M32 10 2 - 5 7 Lt 4 - 1 7 10H or Lt L1 L2 L3 - 2M12 - 2M23 2M12 Lt 6 8 10 10H Chapter 13 Solution 2. L L1 L2 L3 2M12 - 2M23 2M31 10 12 8 2x6 - 2x6 -2x4 22H Chapter 13 Solution 3. L1 L2 2M 250 mH 1 L1 L2 - 2M 150 mH 2 Adding 1 and 2 2L1 2L2 400 mH But L1 3L2 or 8L2 400 and L2 50 mH L1 3L2 150 mH From 2 150 50 - 2M 150 leads to M 25 mH k M a L1L2 2.5 50x150 0.2887 Chapter 13 Solution 4. a For the series connection shown in Figure a the current I enters each coil from its dotted terminal. Therefore the mutually induced voltages have the same sign as the self-induced voltages. Thus Leq L1 L2 2M b Is I1 I2 and Zeq Vs Is Applying KVL to each branch gives Vs jœLiIi joMI2 1 Vs joMIi jo L2I2 2 or 1 _ r joLi joM r m k J _ joM joL2 j k J A -o2LiL2 o2M2 Ai joVs L2 - M A2 joVs Li - M I1 Ai A and I2 A2 A Is Ii I2 Ai A2 A jo Li L2 - 2M Vs -o2 LiL2 - M Zeq Vs Is jo LiL2 - M jo Li L2 - 2M joLeq i.e. Leq L1L2 - M Ll L2 - 2M Chapter 13 Solution 5. a If the coils are connected in series L L1 L2 2M 25 60 2 0.5 25x60 123.7 mH b If they are connected in parallel T L1L2 - M2 25x60 -19.362 L ----------- ------------------mH 24.31 mH L1 L2 - 2M 25 60 - 2x19.36 ---------- Chapter 13 Solution 6. V1 Ri jo L i l i - jo MI2 V2 -joMI R2 joLJL Chapter 13 Solution 7. Applying KVL to the loop 20Z30 I -j6 j8 j12 10 - j4x2 I 10 j6 where I is the loop current. I 20Z30 10 j6 Vo I j12 10 - j4 I 10 j8 20Z30 10 j8 10 j6 22Z37.66 V Chapter 13 Solution 8. Consider the current as shown below. .