1 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 2 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 3 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Mutual inductance M =π√πΏ1 πΏ2 = 0.6√40 × 5 = 8.4853 ππΏ1 = π80 ππΏ2 = π10 ππ = π16.97 π1 = π80πΌ1 − π16.97πΌ2 π2 = −1697πΌ1 + π10πΌ2 Putting the value of π1 πππ πΌ2 π1 + π16.97πΌ2 10 + π16.97 × (−π2) = = .5493∠ − 90π π80 π80 π1 (π‘) = 0.5493 sin ππ‘ π΄ πΌ1 = π2 = −16.97 × (−π0.5493) + π10 × (−π2) = 0 + 9.3216 = 22.0656∠24.99π π£2 (π‘) = 22.065 cos(ππ‘ + 25π ) π Chapter 10: Magnetically Coupled Networks 4 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: 2π» → πππΏ = π8 1π» →= πππΏ = π4 2 = (4 + π8)πΌ1 − π4πΌ2 0 = −π4πΌ1 + (2 + π4)πΌ2 Solving these two equation leads to πΌ2 = 0.2353 − π0.0588 π = 2πΌ2 = 0.4851∠ − 14.046π Thus π£(π‘) = 0.4851 cos(4π‘ − 14.04π ) π Chapter 10: Magnetically Coupled Networks 5 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 6 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: 2π» → πππΏ = π4 0.5π» → πππΏ = π 1 1 πΉ= = −π 2 πππΆ 24 = π4πΌ1 − ππΌ2 0 = −ππΌ1 + (π4 − π)πΌ2 Solving both equation πΌ2 = −π2.1818 ππ = −ππΌ2 = −2.1818 π£π = −2.1818 cos 2π‘ π Chapter 10: Magnetically Coupled Networks 7 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: π=4 ππΆ = −π π = 0.5(1 − π) 1−π ππΏ = πππ = π4 π4π» = π16 π2π» = π8 1||(−π) = − Applying kvl in first loop 12 = (2 + π16)πΌ1 + π4πΌ2 6 = (1 + π8)πΌ1 + π2πΌ2 Applying kvl in second loop (π8 + 0.5 − π0.5)πΌ2 + π4πΌ1 = 0 Solving equations we get πΌ2 = −0.455∠ − 77.41π ππ = πΌ2 (0.5)(1 − π) = 0.3217∠57.59π π£π = 321.7 cos(4π‘ + 57.6π ) ππ Chapter 10: Magnetically Coupled Networks 8 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: ππΏ1 = π200 ππΏ2 = π800 π1 = π1 √πΏ1 πΏ2 = 7π» → πππ = π280 ππΏ3 = π320 ππΏ4 = π720 π2 = 12 π» → πππ = π480 Applying kvl to middle loop π800πΌπ₯ + π320πΌπ₯ + π280πΌ1 − π480πΌ2 = 0 π1120πΌπ₯ + π280 × 5∠0π − π489 × 2∠ − 90π = 0 πΌπ₯ = 1.516∠ − 145.56π π΄ ππ = π320πΌπ₯ − π480πΌ2 π΄ ππ = 794∠ − 150π π π£π = 794 cos(40π‘ − 150π )π Chapter 10: Magnetically Coupled Networks 9 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: π = π (ππΏ − 1 ) ππΆ = π (2 × 106 × 300 × 10−6 − 1012 ) 2 × 106 × 1000 = π100Ω The mutual reactance is ππ = ππ = 120Ω Applying kvl to mesh 1 π(100πΌ1 + 120πΌ2 ) = 10 Applying kvl to mesh 2 π(120πΌ1 + 100πΌ2 ) = 0 −π120 × 10 πΌ2 = = −π0.273 π΄ −1002 + 1202 Chapter 10: Magnetically Coupled Networks 10 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 11 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 12 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 13 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 14 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 15 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 16 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 17 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: For mesh 1 (7 + π6)πΌ1 − (2 + π)πΌ2 = 36∠30π For mesh 2 (6 + π3 − π4)πΌ2 − 2πΌ1 − ππΌ1 = 0 Solving both equation πΌ1 = 4.254∠ − 8.51π , πΌ2 = 1.5637∠27.52π π΄ Chapter 10: Magnetically Coupled Networks 18 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: 800πβ → πππΏ = π480 600ππ» → πππΏ = π360 1200ππ» → πππΏ = π720 1 12ππΉ → = −π138.89 πππΏ For mesh 1 (200 + π480 + π720)πΌ1 + π360πΌ2 − π720πΌ2 = 800 (200 + π1200)πΌ1 − π360πΌ2 = 800 … . . (1) For mesh 2 110∠30π + (150 − π138.89 + π720)πΌ2 + π360πΌ1 = 0 −π360πΌ1 + (150 + π581.1)πΌ2 = −95.2628 − π55 Solving equation (1) and (2) πΌ1 = 0.1390 − π0.7242 πΌ2 = 0.0609 − π0.2690 πΌπ₯ = πΌ1 − πΌ2 = 0.4619∠ − 80.26π Chapter 10: Magnetically Coupled Networks … . (2) 19 Irwin, Engineering Circuit Analysis, 11e ISV ππ₯ = 461.9 cos(600π‘ − 80.26π )ππ΄ Chapter 10: Magnetically Coupled Networks 20 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 21 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 22 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: We can draw following circuit πΌπ = πΌ1 − πΌ3 πΌπ = πΌ2 − πΌ1 πΌπ = πΌ3 − πΌ2 For loop 1 −50 + π20(πΌ3 − πΌ2 )40(πΌ1 − π3 ) + π10(π2 − πΌ1 ) − π30(πΌ3 − πΌ2 ) + π80(πΌ1 − πΌ2 ) − π10(πΌ1 − πΌ2 ) = 0 π100πΌ1 − π60πΌ2 − π40πΌ3 = 50 For loop 2 π10(πΌ1 − πΌ2 ) + π80(πΌ2 − πΌ1 ) + π30(πΌ2 − πΌ1 ) + π60(πΌ2 − πΌ3 ) − π20(πΌ1 − πΌ3 ) + 100πΌ2 = 0 −π60πΌ1 + (100 + π80)πΌ2 − π20πΌ3 = 0 For loop 3 −π50πΌ3 + π20(πΌ1 − πΌ3 ) + π60(πΌ3 − πΌ2 ) + π30(πΌ2 − πΌ1 ) − π10(πΌ2 − πΌ1 ) + π40(πΌ3 − πΌ1 ) − π20(πΌ3 − πΌ2 ) = 0 −π40πΌ1 − π20πΌ2 + π10πΌ3 = 0 Chapter 10: Magnetically Coupled Networks 23 Irwin, Engineering Circuit Analysis, 11e ISV Solving the above equation we get πΌ2 = 0.2355∠42.3π πΌ3 = πΌπ = 1.3049∠63π π΄ Chapter 10: Magnetically Coupled Networks 24 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 25 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: For mesh 1 (10 + π4πΌ1 + π2πΌ2 = 16 For mesh 2 π2πΌ1 + (30 + π26)πΌ2 − π12πΌ3 = 0 For mesh 3 −π12πΌ2 + (5 + π11)πΌ3 = 0 Solving the above equation πΌ1 = 1.3736 − π0.5385 = 1.4754∠ − 21.41π π΄ πΌ2 = −0.0547 − π0.0549 = 0.0775∠ − 134.85π π΄ πΌ3 = −0.0268 − π0.0721 = 0.077∠ − 110.41π π΄ Chapter 10: Magnetically Coupled Networks 26 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: We insert a 1V source at the input For the loop 1 1 = (1 + π10)πΌ1 − π4πΌ2 For loop 2 (8 + π4 + π10 − π2)πΌ2 + π2πΌ1 − π6πΌ1 = 0 ππΌ1 + (2 + π3)πΌ2 = 0 Solving both equation πΌ1 = 0.019 − π0.1068 π= 1 = 1.6154 + π9.077 Ω πΌ1 Chapter 10: Magnetically Coupled Networks 27 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 28 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 29 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 30 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: (See Next Page) Chapter 10: Magnetically Coupled Networks 31 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 32 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 33 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 34 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Using the concept of reflected impedance πππ = π40 + 25 + π30 + = 25 + π70 + Chapter 10: Magnetically Coupled Networks (10)2 8 + π20 − π6 100 = 28.08 + π64.62 8 + π14 35 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 36 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 37 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 38 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 39 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: We find ππ‘β by replacing 20 Ω with 1 V source For mesh 1 (8 − ππ + π½12)πΌ1 − π10πΌ2 = 0 For mesh 2 1 + π15πΌ2 − π10πΌ1 = 0 Solving both equations we get −1.2 + π0.8 + 0.1π 12 + π8 − π1.5π 1 12 + π8 − π1.5π = = −πΌ2 1.2 − π0.8 − 0.1π πΌ2 = ππ‘β |ππ‘β | = 20 = π = 6.425 Chapter 10: Magnetically Coupled Networks √122 + (8 − 1.5π)2 √(1.2 − π0.1π)2 + 0.82 40 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: 30 ππ» → πππΏ = π30 50ππ» → πππΏ = π50 Let π = ππ πππ = 10 + π30 + πΌ1 = π2 20 + π50 165 π = πππ 10 + π30 + π2 20+π50 π = 0.5πΌ12 × 10 = 320 |πΌ1 |2 = 64 → πΌ1 = 8 165(20 + π½50) =8 + (10 + π½30)(20 + π½50) π = 33.86 ππ 38.13 If π = 38.127 → ππ → π = 38.127 ππ» π π= = 0.984 √πΏ1 πΏ2 π2 Applying kvl in first loop 165 = (10 + π30)πΌ1 − π38.127πΌ2 0 = (20 + π50)πΌ2 − π38.127πΌ1 Solving the above equation we get πΌ1 = 8∠ − 13.81π πΌ2 = 5.664∠7.97π Chapter 10: Magnetically Coupled Networks 41 Irwin, Engineering Circuit Analysis, 11e ISV π1 = 8 cos(85.94π + 7.97π ) = 2.457 π2 = 5.664 cos(85.94π + 7.97π ) = −0.3862 π€ = 0.5πΏ1 π12 + 0.5πΏ2 π22 + ππ2 π2 = 130.51 ππ½ Chapter 10: Magnetically Coupled Networks 42 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: (See Next Page) Chapter 10: Magnetically Coupled Networks 43 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 44 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Consider the circuit below We reflect the 200 Ω load to the primary side. 200 = 108 52 10 πΌ1 2 πΌ1 = , πΌ2 = = 108 π 108 1 1 2 2 π = |πΌ2 |2 π πΏ = ( ) (200) = 34.3ππ 2 2 108 ππ = 100 + Chapter 10: Magnetically Coupled Networks 45 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: High voltage side 600 2 ) (0.8∠10π ) = 20∠10π 120 πππ = 60∠ − 30π + 20∠10π = 76.4122∠ − 20.31π ππΏ = ( 600 600 = = 7.8521∠20.31π π΄ πππ 76.4122∠ − 20.31π πΌ1 π£1 π = πΌ1 π£1 = πΌ2 π£2 , πΌ2 = = 39.2605∠20.31π π΄ π£2 πΌ1 = Chapter 10: Magnetically Coupled Networks 46 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Let π1 = π1′ + π1 ′′ Where single prime is due to the dc source and double prime is due to the AC source. Since we are looking for the steady state value π1′ = π2′′ = 0 For AC source π£2 = −π, π£1 πΌ2′′ 1 =− π πΌ1′′ π£π π£2 = π£π , π£1 = − π π£π ′′ πΌ1 = π π πΌ1′′ π£π ′′ πΌ2 = − = − 2 π π π So π1 (π‘) = π£π π£π cos ππ‘ πππ π2 (π‘) = − 2 cos ππ‘ π π π π Chapter 10: Magnetically Coupled Networks 47 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Applying kvl in mesh 1 (50 − π2)πΌ1 + π1 = 80 Mesh 2 −π2 + (2 − π20)πΌ2 = 0 At transformer terminal π2 = 2π1 πΌ1 = 2πΌ2 Solving the equations πΌ2 = 0.8051 − π0.0488 = 0.8056∠ − 347π π = |πΌ2 |2 π = (0.8056)2 × 2 = 1.3012 π Chapter 10: Magnetically Coupled Networks 48 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 49 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 50 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: (See Next Page) Chapter 10: Magnetically Coupled Networks 51 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 52 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 53 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Let π1 = π1′ + π1 ′′ Where single prime is due to the dc source and double prime is due to the AC source. Since we are looking for the steady state value π1′ = π2′′ = 0 For AC source π£2 = −π, π£1 πΌ2′′ 1 ′′ = − π πΌ1 π£π π£2 = π£π , π£1 = − π π£π ′′ πΌ1 = π π πΌ1′′ π£π πΌ2′′ = − = − 2 π π π So π1 (π‘) = π£π π£π cos ππ‘ πππ π2 (π‘) = − 2 cos ππ‘ π π π π Chapter 10: Magnetically Coupled Networks 54 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 55 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 56 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 57 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 58 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 59 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 60 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 61 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 62 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 63 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 64 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 65 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 66 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 67 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 68 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 69 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: For maximum power transfer ππβ = ππΏ π2 π=√ ππΏ = 0.25 ππβ Chapter 10: Magnetically Coupled Networks 70 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: (See Next Page) Chapter 10: Magnetically Coupled Networks 71 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 72 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 73 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 74 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 75 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 76 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: At node 1 (200 − π1 ) π1 − π4 = + πΌ1 10 40 1.25π1 − 0.25π4 + 10πΌ1 At node 2 π1 − π2 π4 = + πΌ3 40 20 3π4 + 40πΌ3 = π1 At the terminals of the first transformer π2 = −2 π1 πΌ2 1 =− πΌ1 2 For the middle loop applying kvl −π2 + 50πΌ2 + π3 = 0 π2 − 50πΌ2 = π3 Chapter 10: Magnetically Coupled Networks 77 Irwin, Engineering Circuit Analysis, 11e ISV For the terminals of the second transformer π4 =3 π3 πΌ3 1 =− πΌ2 3 Solving the above seven equations forπ4 We get π4 = 14.87 π42 π= = 11.05 π 20 Chapter 10: Magnetically Coupled Networks 78 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 79 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 10: Magnetically Coupled Networks 80 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 10: Magnetically Coupled Networks 81 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: (i) For an input of 110 V, winding of the primary coil must be connected in parallel with series aiding on the secondary. Te coils must be series opposing to give 14 V . Thus the connections are as following (ii) To get 220 on the secondary side the primary side the coils are connected in series with series aiding on the secondary side. The coils must be connected series aiding to give 50 V. Thus , the connections are as following Chapter 10: Magnetically Coupled Networks 82 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: ππ 120 1 = = ππ 7200 60 120 1200 πΌπ = 10 × = 144 144 ππ πΌπ πΌπ = = 139 ππ΄ ππ π= Chapter 10: Magnetically Coupled Networks 83 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: For this design we have to consider step down transformer because we need to work on lower level voltage So π1 = 240 π π2 = 120 π So turn ratio is π= π2 120 = = 0.5 π1 240 So for designing the transformer we need to have ratio of secondary winding to primary winding is 0.5. Chapter 10: Magnetically Coupled Networks