ch10

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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 =π√πΏ1 πΏ2 = 0.6√40 &times; 5 = 8.4853
ππΏ1 = π80
ππΏ2 = π10
ππ = π16.97
π1 = π80πΌ1 − π16.97πΌ2
π2 = −1697πΌ1 + π10πΌ2
Putting the value of π1 πππ πΌ2
π1 + π16.97πΌ2 10 + π16.97 &times; (−π2)
=
= .5493∠ − 90π
π80
π80
π1 (π‘) = 0.5493 sin ππ‘ π΄
πΌ1 =
π2 = −16.97 &times; (−π0.5493) + π10 &times; (−π2) = 0 + 9.3216 = 22.0656∠24.99π
π£2 (π‘) = 22.065 cos(ππ‘ + 25π ) π
Chapter 10: Magnetically Coupled Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
2π» → πππΏ = π8
1π» →= πππΏ = π4
2 = (4 + π8)πΌ1 − π4πΌ2
0 = −π4πΌ1 + (2 + π4)πΌ2
Solving these two equation leads to
πΌ2 = 0.2353 − π0.0588
π = 2πΌ2 = 0.4851∠ − 14.046π
Thus π£(π‘) = 0.4851 cos(4π‘ − 14.04π ) π
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:
2π» → πππΏ = π4
0.5π» → πππΏ = π
1
1
πΉ=
= −π
2
πππΆ
24 = π4πΌ1 − ππΌ2
0 = −ππΌ1 + (π4 − π)πΌ2
Solving both equation
πΌ2 = −π2.1818
ππ = −ππΌ2 = −2.1818
π£π = −2.1818 cos 2π‘ π
Chapter 10: Magnetically Coupled Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
π=4
ππΆ = −π
π
= 0.5(1 − π)
1−π
ππΏ = πππ = π4
π4π» = π16
π2π» = π8
1||(−π) = −
Applying kvl in first loop
12 = (2 + π16)πΌ1 + π4πΌ2
6 = (1 + π8)πΌ1 + π2πΌ2
Applying kvl in second loop
(π8 + 0.5 − π0.5)πΌ2 + π4πΌ1 = 0
Solving equations we get
πΌ2 = −0.455∠ − 77.41π
ππ = πΌ2 (0.5)(1 − π) = 0.3217∠57.59π
π£π = 321.7 cos(4π‘ + 57.6π ) ππ
Chapter 10: Magnetically Coupled Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
ππΏ1 = π200
ππΏ2 = π800
π1 = π1 √πΏ1 πΏ2 = 7π» → πππ = π280
ππΏ3 = π320
ππΏ4 = π720
π2 = 12 π» → πππ = π480
Applying kvl to middle loop
π800πΌπ₯ + π320πΌπ₯ + π280πΌ1 − π480πΌ2 = 0
π1120πΌπ₯ + π280 &times; 5∠0π − π489 &times; 2∠ − 90π = 0
πΌπ₯ = 1.516∠ − 145.56π π΄
ππ = π320πΌπ₯ − π480πΌ2 π΄
ππ = 794∠ − 150π π
π£π = 794 cos(40π‘ − 150π )π
Chapter 10: Magnetically Coupled Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
π = π (ππΏ −
1
)
ππΆ
= π (2 &times; 106 &times; 300 &times; 10−6 −
1012
)
2 &times; 106 &times; 1000
= π100Ω
The mutual reactance is
ππ = ππ = 120Ω
Applying kvl to mesh 1
π(100πΌ1 + 120πΌ2 ) = 10
Applying kvl to mesh 2
π(120πΌ1 + 100πΌ2 ) = 0
−π120 &times; 10
πΌ2 =
= −π0.273 π΄
−1002 + 1202
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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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
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:
Chapter 10: Magnetically Coupled Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 10: Magnetically Coupled Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
For mesh 1
(7 + π6)πΌ1 − (2 + π)πΌ2 = 36∠30π
For mesh 2
(6 + π3 − π4)πΌ2 − 2πΌ1 − ππΌ1 = 0
Solving both equation
πΌ1 = 4.254∠ − 8.51π , πΌ2 = 1.5637∠27.52π π΄
Chapter 10: Magnetically Coupled Networks
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SOLUTION:
800πβ → πππΏ = π480
600ππ» → πππΏ = π360
1200ππ» → πππΏ = π720
1
12ππΉ →
= −π138.89
πππΏ
For mesh 1
(200 + π480 + π720)πΌ1 + π360πΌ2 − π720πΌ2 = 800
(200 + π1200)πΌ1 − π360πΌ2 = 800
… . . (1)
For mesh 2
110∠30π + (150 − π138.89 + π720)πΌ2 + π360πΌ1 = 0
−π360πΌ1 + (150 + π581.1)πΌ2 = −95.2628 − π55
Solving equation (1) and (2)
πΌ1 = 0.1390 − π0.7242
πΌ2 = 0.0609 − π0.2690
πΌπ₯ = πΌ1 − πΌ2 = 0.4619∠ − 80.26π
Chapter 10: Magnetically Coupled Networks
… . (2)
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Irwin, Engineering Circuit Analysis, 11e ISV
ππ₯ = 461.9 cos(600π‘ − 80.26π )ππ΄
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:
Chapter 10: Magnetically Coupled Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
We can draw following circuit
πΌπ = πΌ1 − πΌ3
πΌπ = πΌ2 − πΌ1
πΌπ = πΌ3 − πΌ2
For loop 1
−50 + π20(πΌ3 − πΌ2 )40(πΌ1 − π3 ) + π10(π2 − πΌ1 ) − π30(πΌ3 − πΌ2 ) + π80(πΌ1 − πΌ2 ) − π10(πΌ1 − πΌ2 ) = 0
π100πΌ1 − π60πΌ2 − π40πΌ3 = 50
For loop 2
π10(πΌ1 − πΌ2 ) + π80(πΌ2 − πΌ1 ) + π30(πΌ2 − πΌ1 ) + π60(πΌ2 − πΌ3 ) − π20(πΌ1 − πΌ3 ) + 100πΌ2 = 0
−π60πΌ1 + (100 + π80)πΌ2 − π20πΌ3 = 0
For loop 3
−π50πΌ3 + π20(πΌ1 − πΌ3 ) + π60(πΌ3 − πΌ2 ) + π30(πΌ2 − πΌ1 ) − π10(πΌ2 − πΌ1 ) + π40(πΌ3 − πΌ1 ) − π20(πΌ3 − πΌ2 ) = 0
−π40πΌ1 − π20πΌ2 + π10πΌ3 = 0
Chapter 10: Magnetically Coupled Networks
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Solving the above equation we get
πΌ2 = 0.2355∠42.3π
πΌ3 = πΌπ = 1.3049∠63π π΄
Chapter 10: Magnetically Coupled Networks
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SOLUTION:
Chapter 10: Magnetically Coupled Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
For mesh 1
(10 + π4πΌ1 + π2πΌ2 = 16
For mesh 2
π2πΌ1 + (30 + π26)πΌ2 − π12πΌ3 = 0
For mesh 3
−π12πΌ2 + (5 + π11)πΌ3 = 0
Solving the above equation
πΌ1 = 1.3736 − π0.5385 = 1.4754∠ − 21.41π π΄
πΌ2 = −0.0547 − π0.0549 = 0.0775∠ − 134.85π π΄
πΌ3 = −0.0268 − π0.0721 = 0.077∠ − 110.41π π΄
Chapter 10: Magnetically Coupled Networks
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SOLUTION:
We insert a 1V source at the input
For the loop 1
1 = (1 + π10)πΌ1 − π4πΌ2
For loop 2
(8 + π4 + π10 − π2)πΌ2 + π2πΌ1 − π6πΌ1 = 0
ππΌ1 + (2 + π3)πΌ2 = 0
Solving both equation
πΌ1 = 0.019 − π0.1068
π=
1
= 1.6154 + π9.077 Ω
πΌ1
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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Chapter 10: Magnetically Coupled Networks
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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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Chapter 10: Magnetically Coupled Networks
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SOLUTION:
Using the concept of reflected impedance
πππ = π40 + 25 + π30 +
= 25 + π70 +
Chapter 10: Magnetically Coupled Networks
(10)2
8 + π20 − π6
100
= 28.08 + π64.62
8 + π14
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Irwin, Engineering Circuit Analysis, 11e ISV
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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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
We find ππ‘β by replacing 20 Ω with 1 V source
For mesh 1
(8 − ππ + π½12)πΌ1 − π10πΌ2 = 0
For mesh 2
1 + π15πΌ2 − π10πΌ1 = 0
Solving both equations we get
−1.2 + π0.8 + 0.1π
12 + π8 − π1.5π
1
12 + π8 − π1.5π
=
=
−πΌ2 1.2 − π0.8 − 0.1π
πΌ2 =
ππ‘β
|ππ‘β | = 20 =
π = 6.425
Chapter 10: Magnetically Coupled Networks
√122 + (8 − 1.5π)2
√(1.2 − π0.1π)2 + 0.82
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
30 ππ» → πππΏ = π30
50ππ» → πππΏ = π50
Let π = ππ
πππ = 10 + π30 +
πΌ1 =
π2
20 + π50
165
π
=
πππ 10 + π30 +
π2
20+π50
π = 0.5πΌ12 &times; 10 = 320
|πΌ1 |2 = 64 → πΌ1 = 8
165(20 + π½50)
=8
+ (10 + π½30)(20 + π½50)
π = 33.86 ππ 38.13
If π = 38.127 → ππ → π = 38.127 ππ»
π
π=
= 0.984
√πΏ1 πΏ2
π2
Applying kvl in first loop
165 = (10 + π30)πΌ1 − π38.127πΌ2
0 = (20 + π50)πΌ2 − π38.127πΌ1
Solving the above equation we get
πΌ1 = 8∠ − 13.81π
πΌ2 = 5.664∠7.97π
Chapter 10: Magnetically Coupled Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
π1 = 8 cos(85.94π + 7.97π ) = 2.457
π2 = 5.664 cos(85.94π + 7.97π ) = −0.3862
π€ = 0.5πΏ1 π12 + 0.5πΏ2 π22 + ππ2 π2
= 130.51 ππ½
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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
= 108
52
10
πΌ1
2
πΌ1 =
,
πΌ2 = =
108
π 108
1
1 2 2
π = |πΌ2 |2 ππΏ = (
) (200) = 34.3ππ
2
2 108
ππ = 100 +
Chapter 10: Magnetically Coupled Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
High voltage side
600 2
) (0.8∠10π ) = 20∠10π
120
πππ = 60∠ − 30π + 20∠10π = 76.4122∠ − 20.31π
ππΏ = (
600
600
=
= 7.8521∠20.31π π΄
πππ
76.4122∠ − 20.31π
πΌ1 π£1
π = πΌ1 π£1 = πΌ2 π£2 , πΌ2 =
= 39.2605∠20.31π π΄
π£2
πΌ1 =
Chapter 10: Magnetically Coupled Networks
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SOLUTION:
Let π1 = π1′ + π1 ′′
Where single prime is due to the dc source and double prime is due to the AC source. Since we are looking for the
π1′ = π2′′ = 0
For AC source
π£2
= −π,
π£1
πΌ2′′
1
=−
π
πΌ1′′
π£π
π£2 = π£π , π£1 = −
π
π£π
′′
πΌ1 =
ππ
πΌ1′′
π£π
′′
πΌ2 = − = − 2
π
ππ
So
π1 (π‘) =
π£π
π£π
cos ππ‘ πππ π2 (π‘) = − 2 cos ππ‘
ππ
ππ
Chapter 10: Magnetically Coupled Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Applying kvl in mesh 1
(50 − π2)πΌ1 + π1 = 80
Mesh 2
−π2 + (2 − π20)πΌ2 = 0
At transformer terminal
π2 = 2π1
πΌ1 = 2πΌ2
Solving the equations
πΌ2 = 0.8051 − π0.0488 = 0.8056∠ − 347π
π = |πΌ2 |2 π = (0.8056)2 &times; 2 = 1.3012 π
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SOLUTION:
Chapter 10: Magnetically Coupled Networks
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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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Chapter 10: Magnetically Coupled Networks
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SOLUTION:
Let π1 = π1′ + π1 ′′
Where single prime is due to the dc source and double prime is due to the AC source. Since we are looking for the
π1′ = π2′′ = 0
For AC source
π£2
= −π,
π£1
πΌ2′′
1
′′ = − π
πΌ1
π£π
π£2 = π£π , π£1 = −
π
π£π
′′
πΌ1 =
ππ
πΌ1′′
π£π
πΌ2′′ = − = − 2
π
ππ
So
π1 (π‘) =
π£π
π£π
cos ππ‘ πππ π2 (π‘) = − 2 cos ππ‘
ππ
ππ
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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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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
For maximum power transfer
ππβ =
ππΏ
π2
π=√
ππΏ
= 0.25
ππβ
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SOLUTION:
(See Next Page)
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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 − π1 ) π1 − π4
=
+ πΌ1
10
40
1.25π1 − 0.25π4 + 10πΌ1
At node 2
π1 − π2 π4
=
+ πΌ3
40
20
3π4 + 40πΌ3 = π1
At the terminals of the first transformer
π2
= −2
π1
πΌ2
1
=−
πΌ1
2
For the middle loop applying kvl
−π2 + 50πΌ2 + π3 = 0
π2 − 50πΌ2 = π3
Chapter 10: Magnetically Coupled Networks
77
Irwin, Engineering Circuit Analysis, 11e ISV
For the terminals of the second transformer
π4
=3
π3
πΌ3
1
=−
πΌ2
3
Solving the above seven equations forπ4
We get
π4 = 14.87
π42
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:
ππ
120
1
=
=
ππ 7200 60
120 1200
πΌπ  = 10 &times;
=
144
144
ππ  πΌπ
πΌπ =
= 139 ππ΄
ππ
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
π1 = 240 π
π2 = 120 π
So turn ratio is
π=
π2 120
=
= 0.5
π1 240
So for designing the transformer we need to have ratio of secondary winding to primary winding is 0.5.
Chapter 10: Magnetically Coupled Networks
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