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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 10: Magnetically Coupled Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 10: Magnetically Coupled Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Mutual inductance M =𝑘√𝐿1 𝐿2 = 0.6√40 × 5 = 8.4853
𝑍𝐿1 = 𝑗80
𝑍𝐿2 = 𝑗10
𝑍𝑀 = 𝑗16.97
𝑉1 = 𝑗80𝐼1 − 𝑗16.97𝐼2
𝑉2 = −1697𝐼1 + 𝑗10𝐼2
Putting the value of 𝑉1 𝑎𝑛𝑑 𝐼2
𝑉1 + 𝑗16.97𝐼2 10 + 𝑗16.97 × (−𝑗2)
𝐼1 = = = .5493∠ − 90𝑜
𝑗80 𝑗80
𝑖1 (𝑡) = 0.5493 sin 𝜔𝑡 𝐴
𝑉2 = −16.97 × (−𝑗0.5493) + 𝑗10 × (−𝑗2) = 0 + 9.3216 = 22.0656∠24.99𝑜
𝑣2 (𝑡) = 22.065 cos(𝜔𝑡 + 25𝑜 ) 𝑉
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SOLUTION:
2𝐻 → 𝑗𝜔𝐿 = 𝑗8
1𝐻 →= 𝑗𝜔𝐿 = 𝑗4
2 = (4 + 𝑗8)𝐼1 − 𝑗4𝐼2
0 = −𝑗4𝐼1 + (2 + 𝑗4)𝐼2
Solving these two equation leads to
𝐼2 = 0.2353 − 𝑗0.0588
𝑉 = 2𝐼2 = 0.4851∠ − 14.046𝑜
Thus 𝑣(𝑡) = 0.4851 cos(4𝑡 − 14.04𝑜 ) 𝑉
Chapter 10: Magnetically Coupled Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 10: Magnetically Coupled Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
2𝐻 → 𝑗𝜔𝐿 = 𝑗4
0.5𝐻 → 𝑗𝜔𝐿 = 𝑗
1 1
𝐹= = −𝑗
2 𝑗𝜔𝐶
24 = 𝑗4𝐼1 − 𝑗𝐼2
0 = −𝑗𝐼1 + (𝑗4 − 𝑗)𝐼2
Solving both equation
𝐼2 = −𝑗2.1818
𝑉𝑜 = −𝑗𝐼2 = −2.1818
𝑣𝑜 = −2.1818 cos 2𝑡 𝑉
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
𝜔=4
𝑍𝐶 = −𝑗
𝑗
1||(−𝑗) = − = 0.5(1 − 𝑗)
1−𝑗
𝑍𝐿 = 𝑗𝜔𝑀 = 𝑗4
𝑍4𝐻 = 𝑗16
𝑍2𝐻 = 𝑗8
Applying kvl in first loop
12 = (2 + 𝑗16)𝐼1 + 𝑗4𝐼2
6 = (1 + 𝑗8)𝐼1 + 𝑗2𝐼2
Applying kvl in second loop
(𝑗8 + 0.5 − 𝑗0.5)𝐼2 + 𝑗4𝐼1 = 0
Solving equations we get
𝐼2 = −0.455∠ − 77.41𝑜
𝑉𝑜 = 𝐼2 (0.5)(1 − 𝑗) = 0.3217∠57.59𝑜
𝑣𝑜 = 321.7 cos(4𝑡 + 57.6𝑜 ) 𝑚𝑉
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SOLUTION:
𝑍𝐿1 = 𝑗200
𝑍𝐿2 = 𝑗800
𝑀1 = 𝑘1 √𝐿1 𝐿2 = 7𝐻 → 𝑗𝜔𝑀 = 𝑗280
𝑍𝐿3 = 𝑗320
𝑍𝐿4 = 𝑗720
𝑀2 = 12 𝐻 → 𝑗𝜔𝑀 = 𝑗480
Applying kvl to middle loop
𝑗800𝐼𝑥 + 𝑗320𝐼𝑥 + 𝑗280𝐼1 − 𝑗480𝐼2 = 0
𝑗1120𝐼𝑥 + 𝑗280 × 5∠0𝑜 − 𝑗489 × 2∠ − 90𝑜 = 0
𝐼𝑥 = 1.516∠ − 145.56𝑜 𝐴
𝑉𝑜 = 𝑗320𝐼𝑥 − 𝑗480𝐼2 𝐴
𝑉𝑜 = 794∠ − 150𝑜 𝑉
𝑣𝑜 = 794 cos(40𝑡 − 150𝑜 )𝑉
Chapter 10: Magnetically Coupled Networks
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SOLUTION:
1
𝑍 = 𝑗 (𝜔𝐿 − )
𝜔𝐶
1012
= 𝑗 (2 × 106 × 300 × 10−6 − )
2 × 106 × 1000
= 𝑗100Ω
The mutual reactance is
𝑋𝑀 = 𝜔𝑀 = 120Ω
Applying kvl to mesh 1
𝑗(100𝐼1 + 120𝐼2 ) = 10
Applying kvl to mesh 2
𝑗(120𝐼1 + 100𝐼2 ) = 0
−𝑗120 × 10
𝐼2 = = −𝑗0.273 𝐴
−1002 + 1202
Chapter 10: Magnetically Coupled Networks
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SOLUTION:
Chapter 10: Magnetically Coupled Networks
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Chapter 10: Magnetically Coupled Networks
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SOLUTION:
Chapter 10: Magnetically Coupled Networks
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Chapter 10: Magnetically Coupled Networks
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SOLUTION:
Chapter 10: Magnetically Coupled Networks
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SOLUTION:
Chapter 10: Magnetically Coupled Networks
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Chapter 10: Magnetically Coupled Networks
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SOLUTION:
For mesh 1
(7 + 𝑗6)𝐼1 − (2 + 𝑗)𝐼2 = 36∠30𝑜
For mesh 2
(6 + 𝑗3 − 𝑗4)𝐼2 − 2𝐼1 − 𝑗𝐼1 = 0
Solving both equation
𝐼1 = 4.254∠ − 8.51𝑜 , 𝐼2 = 1.5637∠27.52𝑜 𝐴
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SOLUTION:
800𝑚ℎ → 𝑗𝜔𝐿 = 𝑗480
600𝑚𝐻 → 𝑗𝜔𝐿 = 𝑗360
1200𝑚𝐻 → 𝑗𝜔𝐿 = 𝑗720
1
12𝜇𝐹 → = −𝑗138.89
𝑗𝜔𝐿
For mesh 1
(200 + 𝑗480 + 𝑗720)𝐼1 + 𝑗360𝐼2 − 𝑗720𝐼2 = 800
(200 + 𝑗1200)𝐼1 − 𝑗360𝐼2 = 800 … . . (1)
For mesh 2
110∠30𝑜 + (150 − 𝑗138.89 + 𝑗720)𝐼2 + 𝑗360𝐼1 = 0
−𝑗360𝐼1 + (150 + 𝑗581.1)𝐼2 = −95.2628 − 𝑗55 … . (2)
Solving equation (1) and (2)
𝐼1 = 0.1390 − 𝑗0.7242
𝐼2 = 0.0609 − 𝑗0.2690
𝐼𝑥 = 𝐼1 − 𝐼2 = 0.4619∠ − 80.26𝑜
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𝑖𝑥 = 461.9 cos(600𝑡 − 80.26𝑜 )𝑚𝐴
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SOLUTION:
Chapter 10: Magnetically Coupled Networks
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SOLUTION:
Chapter 10: Magnetically Coupled Networks
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SOLUTION:
We can draw following circuit
𝐼𝑎 = 𝐼1 − 𝐼3
𝐼𝑏 = 𝐼2 − 𝐼1
𝐼𝑐 = 𝐼3 − 𝐼2
For loop 1
−50 + 𝑗20(𝐼3 − 𝐼2 )40(𝐼1 − 𝑖3 ) + 𝑗10(𝑖2 − 𝐼1 ) − 𝑗30(𝐼3 − 𝐼2 ) + 𝑗80(𝐼1 − 𝐼2 ) − 𝑗10(𝐼1 − 𝐼2 ) = 0
𝑗100𝐼1 − 𝑗60𝐼2 − 𝑗40𝐼3 = 50
For loop 2
𝑗10(𝐼1 − 𝐼2 ) + 𝑗80(𝐼2 − 𝐼1 ) + 𝑗30(𝐼2 − 𝐼1 ) + 𝑗60(𝐼2 − 𝐼3 ) − 𝑗20(𝐼1 − 𝐼3 ) + 100𝐼2 = 0
−𝑗60𝐼1 + (100 + 𝑗80)𝐼2 − 𝑗20𝐼3 = 0
For loop 3
−𝑗50𝐼3 + 𝑗20(𝐼1 − 𝐼3 ) + 𝑗60(𝐼3 − 𝐼2 ) + 𝑗30(𝐼2 − 𝐼1 ) − 𝑗10(𝐼2 − 𝐼1 ) + 𝑗40(𝐼3 − 𝐼1 ) − 𝑗20(𝐼3 − 𝐼2 ) = 0
−𝑗40𝐼1 − 𝑗20𝐼2 + 𝑗10𝐼3 = 0
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Solving the above equation we get
𝐼2 = 0.2355∠42.3𝑜
𝐼3 = 𝐼𝑜 = 1.3049∠63𝑜 𝐴
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SOLUTION:
Chapter 10: Magnetically Coupled Networks
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SOLUTION:
For mesh 1
(10 + 𝑗4𝐼1 + 𝑗2𝐼2 = 16
For mesh 2
𝑗2𝐼1 + (30 + 𝑗26)𝐼2 − 𝑗12𝐼3 = 0
For mesh 3
−𝑗12𝐼2 + (5 + 𝑗11)𝐼3 = 0
Solving the above equation
𝐼1 = 1.3736 − 𝑗0.5385 = 1.4754∠ − 21.41𝑜 𝐴
𝐼2 = −0.0547 − 𝑗0.0549 = 0.0775∠ − 134.85𝑜 𝐴
𝐼3 = −0.0268 − 𝑗0.0721 = 0.077∠ − 110.41𝑜 𝐴
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SOLUTION:
We insert a 1V source at the input
For the loop 1
1 = (1 + 𝑗10)𝐼1 − 𝑗4𝐼2
For loop 2
(8 + 𝑗4 + 𝑗10 − 𝑗2)𝐼2 + 𝑗2𝐼1 − 𝑗6𝐼1 = 0
𝑗𝐼1 + (2 + 𝑗3)𝐼2 = 0
Solving both equation
𝐼1 = 0.019 − 𝑗0.1068
1
𝑍= = 1.6154 + 𝑗9.077 Ω
𝐼1
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SOLUTION:
Chapter 10: Magnetically Coupled Networks
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SOLUTION:
(See Next Page)
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Chapter 10: Magnetically Coupled Networks
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SOLUTION:
Using the concept of reflected impedance
(10)2
𝑍𝑖𝑛 = 𝑗40 + 25 + 𝑗30 +
8 + 𝑗20 − 𝑗6
100
= 25 + 𝑗70 + = 28.08 + 𝑗64.62
8 + 𝑗14
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SOLUTION:
Chapter 10: Magnetically Coupled Networks
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SOLUTION:
Chapter 10: Magnetically Coupled Networks
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SOLUTION:
Chapter 10: Magnetically Coupled Networks
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SOLUTION:
Chapter 10: Magnetically Coupled Networks
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SOLUTION:
We find 𝑍𝑡ℎ by replacing 20 Ω with 1 V source
For mesh 1
(8 − 𝑗𝑋 + 𝐽12)𝐼1 − 𝑗10𝐼2 = 0
For mesh 2
1 + 𝑗15𝐼2 − 𝑗10𝐼1 = 0
Solving both equations we get
−1.2 + 𝑗0.8 + 0.1𝑋
𝐼2 =
12 + 𝑗8 − 𝑗1.5𝑋
1 12 + 𝑗8 − 𝑗1.5𝑋
𝑍𝑡ℎ = =
−𝐼2 1.2 − 𝑗0.8 − 0.1𝑋
√122 + (8 − 1.5𝑋)2
|𝑍𝑡ℎ | = 20 =
√(1.2 − 𝑗0.1𝑋)2 + 0.82
𝑋 = 6.425
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SOLUTION:
30 𝑚𝐻 → 𝑗𝜔𝐿 = 𝑗30
50𝑚𝐻 → 𝑗𝜔𝐿 = 𝑗50
Let 𝑋 = 𝜔𝑀
𝑋2
𝑍𝑖𝑛 = 10 + 𝑗30 +
20 + 𝑗50
𝑉 165
𝐼1 = = 𝑋2
𝑍𝑖𝑛 10 + 𝑗30 +
20+𝑗50
𝑝 = 0.5𝐼12 × 10 = 320
|𝐼1 |2 = 64 → 𝐼1 = 8
165(20 + 𝐽50)
=8
+ (10 + 𝐽30)(20 + 𝐽50)
𝑋2
𝑋 = 33.86 𝑂𝑅 38.13
If 𝑋 = 38.127 → 𝜔𝑀 → 𝑀 = 38.127 𝑚𝐻
𝑀
𝑘= = 0.984
√𝐿1 𝐿2
Applying kvl in first loop
165 = (10 + 𝑗30)𝐼1 − 𝑗38.127𝐼2
0 = (20 + 𝑗50)𝐼2 − 𝑗38.127𝐼1
Solving the above equation we get
𝐼1 = 8∠ − 13.81𝑜
𝐼2 = 5.664∠7.97𝑜
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𝑖1 = 8 cos(85.94𝑜 + 7.97𝑜 ) = 2.457
𝑖2 = 5.664 cos(85.94𝑜 + 7.97𝑜 ) = −0.3862
𝑤 = 0.5𝐿1 𝑖12 + 0.5𝐿2 𝑖22 + 𝑀𝑖2 𝑖2
= 130.51 𝑚𝐽
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SOLUTION:
(See Next Page)
Chapter 10: Magnetically Coupled Networks
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Chapter 10: Magnetically Coupled Networks
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SOLUTION:
Consider the circuit below
We reflect the 200 Ω load to the primary side.
200
𝑍𝑝 = 100 + = 108
52
10 𝐼1 2
𝐼1 = , 𝐼2 = =
108 𝑛 108
1 1 2 2
𝑃 = |𝐼2 |2 𝑅𝐿 = ( ) (200) = 34.3𝑚𝑊
2 2 108
Chapter 10: Magnetically Coupled Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
High voltage side
600 2
𝑍𝐿 = ( ) (0.8∠10𝑜 ) = 20∠10𝑜
120
𝑍𝑖𝑛 = 60∠ − 30𝑜 + 20∠10𝑜 = 76.4122∠ − 20.31𝑜
600 600
𝐼1 = = = 7.8521∠20.31𝑜 𝐴
𝑍𝑖𝑛 76.4122∠ − 20.31𝑜
𝐼1 𝑣1
𝑛 = 𝐼1 𝑣1 = 𝐼2 𝑣2 , 𝐼2 = = 39.2605∠20.31𝑜 𝐴
𝑣2
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SOLUTION:
Let 𝑖1 = 𝑖1′ + 𝑖1 ′′
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
𝑖1′ = 𝑖2′′ = 0
For AC source
𝑣2 𝐼2′′ 1
= −𝑛, =−
𝑣1 𝐼1′′ 𝑛
𝑣𝑚
𝑣2 = 𝑣𝑚 , 𝑣1 = −
𝑛
′′
𝑣𝑚
𝐼1 =
𝑅𝑛
′′
𝐼1′′ 𝑣𝑚
𝐼2 = − = − 2
𝑛 𝑅𝑛
So
𝑣𝑚 𝑣𝑚
𝑖1 (𝑡) = cos 𝜔𝑡 𝑎𝑛𝑑 𝑖2 (𝑡) = − 2 cos 𝜔𝑡
𝑅𝑛 𝑅𝑛
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SOLUTION:
Applying kvl in mesh 1
(50 − 𝑗2)𝐼1 + 𝑉1 = 80
Mesh 2
−𝑉2 + (2 − 𝑗20)𝐼2 = 0
At transformer terminal
𝑉2 = 2𝑉1
𝐼1 = 2𝐼2
Solving the equations
𝐼2 = 0.8051 − 𝑗0.0488 = 0.8056∠ − 347𝑜
𝑃 = |𝐼2 |2 𝑅 = (0.8056)2 × 2 = 1.3012 𝑊
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SOLUTION:
Chapter 10: Magnetically Coupled Networks
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SOLUTION:
(See Next Page)
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Chapter 10: Magnetically Coupled Networks
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SOLUTION:
Let 𝑖1 = 𝑖1′ + 𝑖1 ′′
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
𝑖1′ = 𝑖2′′ = 0
For AC source
𝑣2 𝐼2′′ 1
= −𝑛, ′′ = − 𝑛
𝑣1 𝐼1
𝑣𝑚
𝑣2 = 𝑣𝑚 , 𝑣1 = −
𝑛
′′
𝑣𝑚
𝐼1 =
𝑅𝑛
𝐼1′′ 𝑣𝑚
𝐼2′′ = − = − 2
𝑛 𝑅𝑛
So
𝑣𝑚 𝑣𝑚
𝑖1 (𝑡) = cos 𝜔𝑡 𝑎𝑛𝑑 𝑖2 (𝑡) = − 2 cos 𝜔𝑡
𝑅𝑛 𝑅𝑛
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SOLUTION:
Chapter 10: Magnetically Coupled Networks
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Chapter 10: Magnetically Coupled Networks
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SOLUTION:
Chapter 10: Magnetically Coupled Networks
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SOLUTION:
For maximum power transfer
𝑍𝐿
𝑍𝑇ℎ =
𝑛2
𝑍𝐿
𝑛=√ = 0.25
𝑍𝑇ℎ
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SOLUTION:
(See Next Page)
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SOLUTION:
At node 1
(200 − 𝑉1 ) 𝑉1 − 𝑉4
= + 𝐼1
10 40
1.25𝑉1 − 0.25𝑉4 + 10𝐼1
At node 2
𝑉1 − 𝑉2 𝑉4
= + 𝐼3
40 20
3𝑉4 + 40𝐼3 = 𝑉1
At the terminals of the first transformer
𝑉2
= −2
𝑉1
𝐼2 1
=−
𝐼1 2
For the middle loop applying kvl
−𝑉2 + 50𝐼2 + 𝑉3 = 0
𝑉2 − 50𝐼2 = 𝑉3
Chapter 10: Magnetically Coupled Networks
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For the terminals of the second transformer
𝑉4
=3
𝑉3
𝐼3 1
=−
𝐼2 3
Solving the above seven equations for𝑉4
We get
𝑉4 = 14.87
𝑉42
𝑃= = 11.05 𝑊
20
Chapter 10: Magnetically Coupled Networks
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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
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
𝑉𝑠 120 1
𝑛= = =
𝑉𝑝 7200 60
120 1200
𝐼𝑠 = 10 × =
144 144
𝑉𝑠 𝐼𝑠
𝐼𝑝 = = 139 𝑚𝐴
𝑉𝑝
Chapter 10: Magnetically Coupled Networks
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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
𝑉1 = 240 𝑉
𝑉2 = 120 𝑉
So turn ratio is
𝑉2 120
𝑛= = = 0.5
𝑉1 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
