Transcript Mesh Current Method

ECE 221
Electric Circuit Analysis I
Chapter 9
Mesh Current Method
Herbert G. Mayer, PSU
Status 10/15/2015
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Syllabus
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Definition
Circuit1 for Mesh Current
First Solve Via Substitution
Example 4.4
Example 4.5
Conclusion
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Definition
 The Mesh Current Method is an alternative to the
substitution method for computing electric units
in circuits
 It expresses voltages as a function of fictitious
currents along a mesh
 Why have yet another method?
 The Mesh Current Method is simpler due to a
smaller number of equations needed
 But the method is applicable only to planar
circuits
 Mesh Currents are not identical to branch currents
 They are fictitious, thus not always measurable
with an instrument (Amp meter)
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Definition
 A Mesh Current is associated with a
complete path through a mesh, obeying
Kirchhoff’s law
 Valid only in a mesh, i.e. any loop with no
interior loops enclosed
 Valid only in the perimeter of a mesh
 Has a defined direction, indicated via an
arrow, to be used consistently across mesh
 If a basic element is affected by multiple
mesh currents, they all have to be accounted
for in computing currents
 Example:
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Circuit 1 for Mesh Current
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First Solve Via Substitution
 Number of essential nodes is 2
 Number of essential branches is 3
 To compute the currents i1, i2, and i3, using
substitution, there is just one independent
KCL equation
 So we need 2 more independent voltage
equations for a solution by substitution
 When complete, compare the result with the
Mesh Current Method
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First Solve Via Substitution
(1)
KCL:
i2 + i3
=
i1
(2)
KVL:
R1*i1 + R3*i3 - v1
=
0
(3)
KVL:
R2*i2 - R3*i3 + v2
=
0
Solve (1) for i3 and substitute into (2) and (3):
Solve Via Substitution
(2)’
v1 =
i1*(R1 + R3) - i2*R3
(3)’
v2 =
-i2*(R2 + R3) + i1*R3
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Then Solve Via Mesh Current
(1)
R1*ia + R3*(ia – ib) – v1
=
0
(2)
R2*ib + v2 + R3*(ib – ia)
=
0
(1)”
v1
=
ia*(R1 + R3) – ib*R3
(2)”
v2
= -ib*(R2 + R3) + ia*R3
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Substitution vs. Mesh Current
i1 == ia
// == stands for “identical to”
i2 == ib
i3 = ia - ib
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Conclusion
 Mesh Current Method is simpler than
Substitution
 Equations are reduced to a smaller number
of unknown currents
 Similar to Node Voltage Method, which
reduces equations to a smaller number of
unknown voltages
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Example 4.4
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Example 4.4 via Mesh Current
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Example 4.4 via Mesh Current
 Find the power associated with each voltage
source:
 First find the two interesting currents, the
ones through the constant voltage sources
 Their voltages are known, hence their two
currents must be found to compute power
 There are b = 7 branches with unknown
currents
 And n = 5 nodes
 Hence we need b-(n-1) = 7-(5-1) = 3 equations
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Example 4.4 via Mesh Current
(1) 2*ia + 8*(ia - ib) - 40
=
0
(2) 6*ib + 6*(ib - ic) + 8*(ib - ia)
=
0
(3) 4*ic + 20 + 6*(ic - ib)
=
0
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Example 4.4 via Mesh Current
(1) 2*ia + 8*(ia - ib) - 40
=
0
(2) 6*ib + 6*(ib - ic) + 8*(ib - ia)
=
0
(3) 4*ic + 20 + 6*(ic - ib)
=
0
. . .
(1’)
ia =
4 + 8*ib / 10
(3’)
ic = -2 + 6*ib / 10
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Example 4.4 via Mesh Current
(1) 2*ia + 8*(ia - ib) - 40
=
0
(2) 6*ib + 6*(ib - ic) + 8*(ib - ia)
=
0
(3) 4*ic + 20 + 6*(ic - ib)
=
0
. . .
(1’)
ia
=
4 + 8*ib / 10
(3’)
ic
=
-2 + 6*ib / 10
Substitute both (1’) and (3’) in (2)
. . .
ia
=
5.6 A
ib
=
2.0 A
ic
=
-0.8 A
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Example 4.5
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Next: Example 4.5 via Mesh Current
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Example 4.5 via Mesh Current
 Use the Mesh Current Method to compute the
power P4 dissipated in the 4 Ω resistor
 To this end, compute the currents i2, and i3
in the 3 meshes; also find i1
 With 3 unknowns we need 3 equations
 And need to express the branch current that
controls the dependent voltage source as a
function of other currents
 Then we can express the power P4 consumed
in the 4 Ω resistor
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Example 4.5 via Mesh Current
(1)
20*(i1 - i3) - 50
+ 5*(i1 - i2)
=
0
(2)
5*(i2 - i1) + i2
+ 4*(i2 - i3)
=
0
(3)
4*(i3 - i2) + 15*iα + 20*(i3 - i1) =
0
express iα as a function of the other currents:
iα =
i1 - i3
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Example 4.5 via Mesh Current
(1)
20*(i1 - i3) - 50
+ 5*(i1 - i2)
=
0
(2)
5*(i2 - i1) + i2
+ 4*(i2 - i3)
=
0
(3)
4*(i3 - i2) + 15*iα + 20*(i3 - i1) =
0
express iα as a function of the other currents:
iα =
i1 – i3
(1’) 50
=
25*i1 - 5*i2
(2’)
0
=
-5*i1 + 10*i2 -
4*i3
(3’)
0
=
-5*i1 - 4*i2
9*i3
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- 20*i3
+
Example 4.5 via Mesh Current
(1)
20*(i1 - i3) - 50
+ 5*(i1 - i2)
=
0
(2)
5*(i2 - i1) + i2
+ 4*(i2 - i3)
=
0
(3)
4*(i3 - i2) + 15*iα + 20*(i3 - i1) =
0
express iα as a function of the other currents
iα =
i1 – i3
(1’) 50
=
25*i1 - 5*i2
(2’)
0
=
-5*i1 + 10*i2 -
4*i3
(3’)
0
=
-5*i1 - 4*i2
9*i3
i1 = 29.6 A
- 20*i3
+
i2 = 26 A
i3 = 28 A
Current in 4 Ω resistor = 28 A - 26 A = 2 A
P4 =
2*2*4 =
16 W
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