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MAE140 Win12 HW4 Solutions T R & T, 3.9, 3.10, 3.12, 3.15 a) and c), 3.20 a) and b), 3.22, 3.26, 4.2, 4.6, 4.13 3-Ββ9
(a) To write mesh-Ββcurrent equations by inspection, we note that the total resistances in mesh A and B are
10ππΊ + 5ππΊ, and 10ππΊ + 5ππΊ, respectively. The resistance common to meshes A and B is 5ππΊ. Using these observations, we write the mesh equations as,
Mesh A: 10ππΊ + 5ππΊ π! β 5ππΊπ! β 12π = 0
Mesh B: 10ππΊ + 5ππΊ π! β 5ππΊπ! + 4π = 0
(b) Solving the equations, we can get:
π! = 0.8ππ΄, π! = 0ππ΄
(c) Using the results,
π£! = 10ππΊπ! = 8π
π! = π! β π! = 0.8ππ΄
3-Ββ10
(a) Writing mesh-Ββcurrent equations by inspection, we can get:
πππ β π΄: 4ππΊ + 2ππΊ + 4ππΊ π! β 4ππΊπ! β 15π + 15π = 0
πππ β π΅: 4ππΊ + 2ππΊ + 4ππΊ π! β 4ππΊπ! β 15π = 0
(b) Solving the matrix, we can get
π! = 0.7143 ππ΄, π! = 1.7857 ππ΄
(c) Using the results,
π£! = 4ππΊ π! = 7.14π
π! = π! β π! = 1.07 ππ΄
2
3-Ββ12
(a) Creating a supermesh as shown in the figure, we can get
π! = π! β π!
π ! + π ! π! + π ! + π ! + π ! π! β π£! = 0
(b) Using the given values, we can get
π! = β25ππ΄, π! = 75ππ΄
So we obtain,
π£! = π!π ! = 18.75π π! = βπ! = 0.025 π΄
(c) The total power π = π!! π ! + π ! + π!! π ! + π ! + π ! β π!π£! = 3.75π. Please pay attention to the direction of the current and voltage.
3-Ββ15
(a) Writing mesh-Ββcurrent equations by inspection, we can get: πππ β π΄: π! 2ππΊ + 4ππΊ β π! β 4ππΊ β π! β 2ππΊ = 40π πππ β π΅: β π! β 4ππΊ + π! 4ππΊ + 8ππΊ = 25π πππ β πΆ: π! = 5ππ΄
(c) Node voltage; there is only one equation to solve.
π! π!
supermesh
3
3-Ββ20
(a) Writing the mesh-Ββcurrent equations by inspection, we can get: πππ β π΄: 4ππΊ + 10ππΊ π! β 10ππΊ β π! + 10 = 0 πππ β π΅: β 10ππΊπ! + 1ππΊ + 10ππΊ + 2ππΊ π! β 2ππΊπ! = 0 πππ β πΆ: β 2ππΊπ! + 2ππΊ + 8ππΊ + 6ππΊ π! β 6ππΊπ! = 0 πππ β π·: 6ππΊπ! β 6ππΊπ! β 10 β 5 = 0 Write in matrix form: 4ππΊ + 10ππΊ β10ππΊ 0 0β10ππΊ 1ππΊ + 10ππΊ + 2ππΊ β2ππΊ 00 β2ππΊ 2ππΊ + 8ππΊ + 6ππΊ β6ππΊ0 0 β6ππΊ 6ππΊ
π!π!π!π!
=
β100015
Solving the equations, we can get π! = β1.2565 ππ΄ π! = β0.7592 ππ΄ π! = 1.3482 ππ΄ π! = 3.8482 ππ΄
(b) By looking at the figure, we can find v! = 5V, v! = 15V. Then we can write the node-Ββvoltage equations.
Node B: β !!π£! +
!!+ !
!"+ 1 π£! β π£! β
!!"π£! = 0
Node C: βπ£! + 1 + !!+ !
!π£! β
!!π£! = 0
Solving the equations, we can get π£! = 10.0262 V π£! = 10.7853 V Itβs obvious that node-Ββvoltage needs less effort.
3-Ββ22
Using current division, we can get
π! =
1π ! + π !
1π !+ 1π !
+ 1π ! + π !
π! β πΎ =π!π!=
1π ! + π !
1π !+ 1π !
+ 1π ! + π !
A
B
C
D
4
_+
12V
200
200
100
6V
200
200
100
_+
3-Ββ26
π£!! = 12π β !""||!""!""!!""||!""
= 3π π£!! = 6π β !""||!""!""!!""||!""
= 3π
π£! = π£!! + π£!! = 6π 4-Ββ2
Using the current division in the input circuit, we can get
π! =100
100 + 100π! =
12π!
Similarly, the output current π! can be found in the output circuit by current division.
π! =2
2 + 2π! =
12π!
At node A, KCL requires that π! = β100π! = β50π!. So we can get
π! =12π! = β25π! β
π!π!= β25
Using Ohmβs law, we can get π£! = 2000π! = β5Γ10!π!
π£! = 100π! = 50π! π£!π£!
=β5Γ10!π!20π!
= β1000
The power supplied by the independent current source is π! = 100||100 π!! = 0.2 ππ
The power delivered to the 2πΞ© load is π! = 2000π!! = 5π
A π!
5
Applying KCL at node A, we can get:
π! + π½π! β π! = 0 β π! =1
1 β π½π!
π! = βπ½π! =π½
π½ β 1π! β
π!π!=
π½π½ β 1
Using Ohmβs law, we can get
π !" =π£!"π!
=π!π π!
At node A, applying KCL,
π! β π! + π½π! = 0 β π! =1
1 β π½π!
So the result is
π !" =1
1 β π½π
+
π£!"
-Ββ
A