Bài giải phần giải mạch P13

Chapter 13, Solution 1. For coil 1, L 1 – M 12 + M 13 = 6 – 4 + 2 = 4 For coil 2, L 2 – M 21 – M 23 = 8 – 4 – 5 = – 1 For coil 3, L 3 + M 31 – M 32 = 10 + 2 – 5 = 7 L T = 4 – 1 + 7 = 10H or L T = L 1 + L 2 + L 3 – 2M 12 – 2M 23 + 2M 12 L T = 6 + 8 + 10 = 10H Chapter 13, Solution 2. L = L 1 + L 2 + L 3 + 2M 12 – 2M 23 2M 31 = 10 + 12 +8 + 2x6 – 2x6 –2x4 = 22H Chapter 13, Solution 3. L 1 + L 2 + 2M = 250 mH (1) L 1 + L 2 – 2M = 150 mH (2) Adding (1) and (2), 2L 1 + 2L 2 = 400 mH But, L 1 = 3L 2, , or 8L 2 + 400, and L 2 = 50 mH L 1 = 3L 2 = 150 mH From (2), 150 + 50 – 2M = 150 leads to M = 25 mH k = M/ 150x50/5.2LL 21 = = 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, L eq = L 1 + L 2 + 2M L 2 L 1 L 2 L 1 I 1 I 2 I s V s + – L eq (a) (b) (b) For the parallel coil, consider Figure (b). I s = I 1 + I 2 and Z eq = V s /I s Applying KVL to each branch gives, V s = jωL 1 I 1 + jωMI 2 (1) V s = jωMI 1 + jω L 2 I 2 (2) or              ωω ωω =      2 1 2 1 s s I I LjMj MjLj V V ∆ = –ω 2 L 1 L 2 + ω 2 M 2 , ∆ 1 = jωV s (L 2 – M), ∆ 2 = jωV s (L 1 – M) I 1 = ∆ 1 /∆, and I 2 = ∆ 2 /∆ I s = I 1 + I 2 = (∆ 1 + ∆ 2 )/∆ = jω(L 1 + L 2 – 2M)V s /( –ω 2 (L 1 L 2 – M)) Z eq = V s /I s = jω(L 1 L 2 – M)/[jω(L 1 + L 2 – 2M)] = jωL eq i.e., L eq = (L 1 L 2 – M)/(L 1 + L 2 – 2M) Chapter 13, Solution 5. (a) If the coils are connected in series, =++=++= 60x25)5.0(26025M2LLL 21 123.7 mH (b) If they are connected in parallel, = −+ − = −+ − = mH 36.19x26025 36.1960x25 M2LL MLL L 2 21 2 21 24.31 mH Chapter 13, Solution 6. V 1 = (R 1 + jωL 1 )I 1 – jωMI 2 V 2 = –jωMI 1 + (R 2 + jωL 2 )I 2 Chapter 13, Solution 7. Applying KVL to the loop, 20∠30 ° = I(–j6 + j8 + j12 + 10 – j4x2) = I(10 + j6) where I is the loop current. I = 20∠30°/(10 + j6) V o = I(j12 + 10 – j4) = I(10 + j8) = 20∠30°(10 + j8)/(10 + j6) = 22∠37.66° V Chapter 13, Solution 8. Consider the current as shown below. j 6 j 4 1 Ω -j3 j2 10 + – I 2 I 1 4 Ω + V o – For mesh 1, 10 = (1 + j6)I 1 + j2I 2 (1) For mesh 2, 0 = (4 + j4 – j3)I 2 + j2I 1 0 = j2I 1 +(4 + j)I 2 (2) In matrix form,             + + =       2 1 I I j42j 2j6j1 0 10 ∆ = 2 + j25, and ∆ 2 = –j20 I 2 = ∆ 2 /∆ = –j20/(2 + j25) V o = –j3I 2 = –60/(2 + j25) = 2.392∠94.57° Chapter 13, Solution 9. Consider the circuit below. 2 Ω -j1 2 Ω For loop 1, -j2V + – j 4 j 4 8∠30 o + – I 2 I 1 8∠30° = (2 + j4)I 1 – jI 2 (1) For loop 2, ((j4 + 2 – j)I 2 – jI 1 + (–j2) = 0 or I 1 = (3 – j2)i 2 – 2 (2) Substituting (2) into (1), 8∠30° + (2 + j4)2 = (14 + j7)I 2 I 2 = (10.928 + j12)/(14 + j7) = 1.037∠21.12° V x = 2I 2 = 2.074∠21.12° Chapter 13, Solution 10. Consider the circuit below. I o j ω L j ω L 1/j ω C I 2 I 1 j ω M I in ∠0 o 2 21 LLLkM == = L, I 1 = I in ∠0°, I 2 = I o I o (jωL + R + 1/(jωC)) – jωLI in – (1/(jωC))I in = 0 I o = j I in ( ω L – 1/( ω C)) /(R + j ω L + 1/(j ω C)) Chapter 13, Solution 11. Consider the circuit below. I 1 j ω L 1 R 1 R 2 j ω L 2 1/j ω C j ω M V 2 + – V 1 + – I 3 I 2 For mesh 1, V 1 = I 1 (R 1 + 1/(j ω C)) – I 2 (1/j ω C)) –R 1 I 3 For mesh 2, 0 = –I 1 (1/(j ω C)) + (j ω L 1 + j ω L 2 + (1/(j ω C)) – j2 ω M)I 2 – j ω L 1 I 3 + j ω MI 3 For mesh 3, –V 2 = –R 1 I 1 – jω(L 1 – M)I 2 + (R 1 + R 2 + jωL 1 )I 3 or V 2 = R 1 I 1 + j ω (L 1 – M)I 2 – (R 1 + R 2 + j ω L 1 )I 3 Chapter 13, Solution 12. Let .1= ω j4 j2 • + j6 j8 j10 1V - I 1 I 2 • Applying KVL to the loops, 21 481 IjIj += (1) 21 1840 IjIj += (2) Solving (1) and (2) gives I 1 = -j0.1406. Thus H 111.7 11 11 ==→== jI LjL I Z eqeq We can also use the equivalent T-section for the transform to find the equivalent inductance. Chapter 13, Solution 13. We replace the coupled inductance with an equivalent T-section and use series and parallel combinations to calculate Z. Assuming that ,1= ω 10,101020,81018 21 ===−=−==−=−= MLMLLMLL cba The equivalent circuit is shown below: 12 Ω j8 Ω j10 Ω 2 Ω j10 Ω -j6 Ω Z j4 Ω Z =12 +j8 + j14//(2 + j4) = 13.195 + j11.244Ω Chapter 13, Solution 14. To obtain V Th , convert the current source to a voltage source as shown below. 5 Ω -j3 Ω + – 8 V + V Th – 2 Ω j8 Ω a b j10 V + – j6 Ω I j2 Note that the two coils are connected series aiding. ωL = ωL 1 + ωL 2 – 2ωM jωL = j6 + j8 – j4 = j10 Thus, –j10 + (5 + j10 – j3 + 2)I + 8 = 0 I = (– 8 + j10)/ (7 + j7) But, –j10 + (5 + j6)I – j2I + V Th = 0 V Th = j10 – (5 + j4)I = j10 – (5 + j4)(–8 + j10)/(7 + j7) V Th = 5.349∠34.11° To obtain Z Th , we set all the sources to zero and insert a 1-A current source at the terminals a–b as shown below. I 2 I 1 1 A 5 Ω j6 Ω j8 Ω -j3 Ω a j2 2 Ω + V o – b Clearly, we now have only a super mesh to analyze. (5 + j6)I 1 – j2I 2 + (2 + j8 – j3)I 2 – j2I 1 = 0 (5 + j4)I 1 + (2 + j3)I 2 = 0 (1) But, I 2 – I 1 = 1 or I 2 = I 1 – 1 (2) Substituting (2) into (1), (5 + j4)I 1 +(2 + j3)(1 + I 1 ) = 0 I 1 = –(2 + j3)/(7 + j7) Now, ((5 + j6)I 1 – j2I 1 + V o = 0 V o = –(5 + j4)I 1 = (5 + j4)(2 + j3)/(7 + j7) = (–2 + j23)/(7 + j7) = 2.332∠50° Z Th = V o /1 = 2.332∠50° ohms Chapter 13, Solution 15. To obtain I N , short-circuit a–b as shown in Figure (a). I 1 I 2 I 1 20 Ω j10 Ω j20 Ω a j5 1 + – 20 Ω j10 Ω j20 Ω a j5 I N 60∠30 o + – I 2 b b (a) (b) For mesh 1, 60 ∠ 30 ° = (20 + j10)I 1 + j5I 2 – j10I 2 or 12 ∠ 30 ° = (4 + j2)I 1 – jI 2 (1) For mesh 2, 0 = (j20 + j10)I 2 – j5I 1 – j10I 1 or I 1 = 2I 2 (2) Substituting (2) into (1), 12 ∠ 30 ° = (8 + j3)I 2 I N = I 2 = 12 ∠ 30 ° /(8 + j3) = 1.404 ∠ 9.44 ° A To find Z N , we set all the sources to zero and insert a 1-volt voltage source at terminals a– b as shown in Figure (b). For mesh 1, 1 = I 1 (j10 + j20 – j5x2) + j5I 2 1 = j20I 1 + j5I 2 (3) For mesh 2, 0 = (20 + j10)I 2 + j5I 1 – j10I 1 = (4 + j2)I 2 – jI 1 or I 2 = jI 1 /(4 + j2) (4) Substituting (4) into (3), 1 = j20I 1 + j(j5)I 1 /(4 + j2) = (–1 + j20.5)I 1 I 1 = 1/(–1 + j20.5) Z N = 1/I 1 = (–1 + j20.5) ohms Chapter 13, Solution 16. To find I N , we short-circuit a-b. j Ω 8 Ω -j2 Ω a •• + j4 Ω j6 Ω I 2 I N 80 V I o 0 ∠ 1 - b 80)28(0)428(80 2121 =−+→=−+−+− jIIjjIIjj (1) 2112 606 IIjIIj =→=− (2) Solving (1) and (2) leads to A 91.126246.1362.0584.1 1148 80 2 o N j j II −∠=−= + == To find Z N , insert a 1-A current source at terminals a-b. Transforming the current source to voltage source gives the circuit below. j Ω 8 Ω -j2 Ω 2 Ω a •• + j4 Ω j6 Ω I 2 2V I 1 - b 28 )28(0 2 121 j jI IjIIj + =→−+= (3) 0)62(2 12 =−++ jIIj (4) Solving (3) and (4) leads to I 2 = -0.1055 +j0.2975, V ab =-j6 I 2 = 1.7853 +0.6332 Ω∠== o ab N 53.19894.1 1 V Z