Network Analysis V
Mesh Equations – Three Loops
Circuit overview
R1
A
V1
12 V
B
R4
C
R7
E
2Ω
2Ω
2Ω
IA
IB
IC
R2
2Ω
R3
2Ω
D
V2
8V
R5
2Ω
R6
2Ω
G
R8
F
2Ω
H
Using the method
of mesh currents,
solve for all the
unknown values
of voltage and
current in the
figure shown. To
do this, we will
complete steps a
through p.
Listing of steps
a. Identify the components through
which the mesh current IA flows.
b. Identify the components through
which the mesh current IB flows.
c. Identify the components through
which the mesh current IC flows.
Listing of steps (cont.)
d. Which components have opposing
mesh currents (if any)?
e. Write the mesh equation for mesh A.
f. Write the mesh equation for mesh B.
g. Write the mesh equation for mesh C.
Listing of steps (cont.)
h. Solve for the currents IA, IB and IC using any
of the methods for the solution of
simultaneous equations.
i. Determine the values of currents I1, I2, I3,
I4, I5, I6, I7 and I8.
j. Are the assumed directions of the three
mesh (mesh A, mesh B and mesh C)
currents correct? How do you know?
Listing of steps (cont.)
k. What are the directions of current I2 and I5
through the components R2 and R5
respectively?
l. Solve for the voltage drops VR1, VR2, VR3, VR4,
VR5, VR6, VR7 and VR8.
m. Using the final solutions for VR1, VR2 and VR3,
write a KVL equation for the loop ACDBA
going clockwise from point A.
Listing of steps (cont.)
n. Using the final solutions for VR2, VR4, VR5 and
VR6, write a KVL equation for the loop EFDCE
going clockwise from point E.
o. Using the final solutions for VR5, VR7 and VR8,
write a KVL equation for the loop GHFEG
going clockwise from point G.
p. Using the final solutions for VR1, VR3, VR4, VR6,
VR7 and VR8, write a KVL equation for the loop
ACEGHFDBA going clockwise from point A.
Step a solution
R1
A
C
2Ω
+
_
V1
12 V
IA
R2
2Ω
R3
B
2Ω
a. Identify the
components through
which the mesh
current IA flows.
D
• IA flows through V1, R1,
R2 and R3 (not
necessarily in that
order).
Step b solution
C
R4
E
2Ω
IB
R2
2Ω
R5
2Ω
R6
D
2Ω
F
b. Identify the
components through
which the mesh
current IB flows.
• IB flows through R2, R4,
R5 and R6 (not
necessarily in that
order).
Step c solution
E
R7
G
2Ω
IC
R5
2Ω
+
V2
8V
_
R8
F
2Ω
H
c. Identify the
components through
which the mesh
current IC flows.
• IC flows through V2, R5,
R7 and R8 (not
necessarily in that
order)
Step d solution
d. Which components have
opposing mesh currents
(if any)?
R1
A
R4
C
2Ω
V1
12 V
B
IA
R2
2Ω
R3
2Ω
2Ω
2Ω
IB
IC
R5
2Ω
R6
D
R7
E
2Ω
R8
F
2Ω
G
• The components with
the opposing mesh
currents are R2 and R5
(down through R2 from IA
and up through R2 from
IB, while for R5 it is down
through it from IB and up
through it from IC).
V2
8V
H
Step e solution
R1
A
C
2Ω
+
_
V1
12 V
IA
R2
2Ω
R3
B
2Ω
D
e. Write the mesh
equation for mesh A.
• 6IA – 2IB + 0IC = 12V
Step f solution
C
R4
E
f. Write the mesh
equation for mesh B.
2Ω
IB
R2
2Ω
R5
2Ω
R6
D
2Ω
F
• -2IA + 8IB – 2IC = 0V
Step g solution
E
R7
G
g. Write the mesh
equation for mesh C.
2Ω
IC
R5
2Ω
+
V2
8V
_
R8
F
2Ω
H
• 0IA – 2IB + 6IC = -8V
Step h solution
Original equations:
• Step e:
6IA – 2IB + 0 = 12V
• Step f:
-2IA + 8IB – 2IC = 0V
• Step g:
0 – 2IB + 6IC = -8V
h. Solve for the currents IA, IB
and IC using any of the
methods for the solution
of simultaneous
equations.
• Step 1  Since equation
from step f is equal to zero,
we will rewrite it to solve
for IC
-2IA + 8IB = 2IC (reorder
as 2IC = -2IA + 8IB)
Step h solution (cont.)
• Step 2 Divide by 2 to reduce
𝑰𝑪 = −𝑰𝑨 + 𝟒𝑰𝑩
• Step 3  Substitute into equation from step g
to eliminate third variable (IC)
𝟎 − 𝟐𝑰𝑩 + 𝟔 −𝑰𝑨 + 𝟒𝑰𝑩 = −𝟖
Step h solution (cont.)
• Step 4 Multiply through and combine like
terms
𝟎 − 𝟐𝑰𝑩 − 𝟔𝑰𝑨 + 𝟐𝟒𝑰𝑩 = −𝟖
𝟎 − 𝟔𝑰𝑨 + 𝟐𝟐𝑰𝑩 = −𝟖
Step h solution (cont.)
Step 5  Examine remaining expression. Convert
how necessary to eliminate one term and then add
(Remaining expression: 6𝐼𝐴 − 2𝐼𝐵 = 12; Since this
expression contains the value of IA necessary to eliminate IA,
nothing more needs to be done except add)
𝟔𝑰𝑨 − 𝟐𝑰𝑩 = 𝟏𝟐
−𝟔𝑰𝑨 + 𝟐𝟐𝑰𝑩 = −𝟖
_____________________________
𝟐𝟎𝑰𝑩 = 𝟒
Step h solution (cont.)
• Step 6 Solve for IB by dividing multiplier
𝟐𝟎𝑰𝑩
𝟒
=
∴ 𝑰𝑩 = 𝟎. 𝟐𝐀
𝟐𝟎
𝟐𝟎
Step h solution (cont.)
• Step 7  Substitute IB into either equation to
solve for IA (we will use one from step e as it is
the easiest)
6IA – 2(0.2A) = 12  6IA – 0.4 = 12
Step h solution (cont.)
• Step 7a Move all the constants to the right
side, which will leave our unknown on the left
6IA= 12.4
• Step 7b  Divide multiplier to solve for IA
𝟔𝑰𝑨
𝟏𝟐. 𝟒
=
∴ 𝑰𝑨 = 𝟐. 𝟎𝟔𝟕𝐀
𝟔
𝟔
Step h solution (cont.)
• Step 8  Substitute IB into equation from step
g to solve for IC
-2(0.2A) + 6IC = -8 -0.4 + 6IC = -8
Step h solution (cont.)
• Step 8a Move all constants to the right
which once again leaves our unknown on the
left
6IC= -7.6
• Step 8b  Divide multiplier to solve for IC
𝟔𝑰𝑪
−𝟕. 𝟔
=
∴ 𝑰𝑪 = −𝟏. 𝟐𝟔𝟕𝐀
𝟔
𝟔
Step i solution
i. Determine the values of currents I1, I2, I3, I4,
I5, I6, I7 and I8.
R1
A
R4
C
2Ω
•
•
•
•
•
2Ω
V1
12 V
R3
2Ω
D
V2
8V
R5
2Ω
R6
2Ω
G
2Ω
R2
2Ω
I1 = I3 = IA = 2.067A
I4 = I6 = IB = 0.2A
I7 = I8 = IC = -1.267A
I2 = IA – IB = 2.067A – 0.2A = 1.867A
I5 = IC – IB = -1.267A – 0.2A = -1.467A
B
R7
E
R8
F
2Ω
H
Step j solution
j. Are the assumed directions of the three
mesh (mesh A, mesh B and mesh C) currents
correct? How do you know?
• Following conventional current flow, the
meshes of A and B were correct, but the
direction of mesh C needs to be reversed
(counterclockwise) since it was negative.
Step k solution
k. What are the directions of current I2 and I5
through the components R2 and R5
respectively?
• Current is flowing down through R2 (IA and IB
are subtractive). The current also flows
downward through R5 (IB and IC are additive).
Step l solution
l. Solve for the voltage drops VR1, VR2, VR3, VR4,
VR5, VR6, VR7 and VR8.
•
•
•
•
VR1 = I1R1 = 2.067A(2Ω) = 4.134V
VR2 = I2R2 = 1.867A(2Ω) = 3.734V
VR3 = I3R3 = 2.067A(2Ω) = 4.134V
VR4 = I4R4 = 0.2A(2Ω) = 0.4V
Step l solution (cont.)
l. Solving for the voltage drops VR1, VR2, VR3,
VR4, VR5, VR6, VR7 and VR8 (cont.).
•
•
•
•
VR5 = I5R5 = -1.467A(2Ω) = -2.934V
VR6 = I6R6 = 0.2A(2Ω) = 0.4V
VR7 = I7R7 = -1.267A(2Ω) = -2.534V
VR8 = I8R8 = -1.267A(2Ω) = -2.534V
Step m solution
A
+
R1 _
C
4.134V
+
_
+
V1
12 V
R2
m. Using the final solutions
for VR1, VR2 and VR3, write
a KVL equation for the
loop ACDBA going
clockwise from point A.
3.734V
_
4.134V + 3.734V + 4.134V – 12V = 0
R3
B
_
4.134V
+
D
Step n solution
C
+
R4
_
E
0.4V
+
+
3.734V
R2
_
R5
n. Using the final solutions
for VR2, VR4, VR5 and VR6,
write a KVL equation for
the loop EFDCE going
clockwise from point E.
2.934V
_
2.934V + 0.4V – 3.734V + 0.4V = 0
R6
D
_
0.4V
+
F
Step o solution
_ R7
E
+
G
2.534V
+
V2
8V
_
+
2.934V
R5
_
o. Using the final solutions
for VR5, VR7 and VR8,
write a KVL equation for
the loop GHFEG going
clockwise from point G.
8V – 2.534V – 2.934V – 2.534V = 0
+
F
R8 _
H
2.534V
Step p solution
p. Using the final solutions for VR1, VR3, VR4, VR6,
VR7 and VR8, write a KVL equation for the loop
ACEGHFDBA going clockwise from point A.
4.134V + 0.4V – 2.534V + 8V – 2.354V + 0.4V + 4.134V – 12V = 0
A
+
R1
_
C
4.134V
_
+ R4
_ R7
E
0.4V
G
2.534V
+
+
V1
12 V
V2
8V
_
_
B
+
_ R3
4.134V
_ R6
+
D
0.4V
+
+
F
R8 _
2.534V
H
The circuit with voltage measurements
according to original current assumption
U1
+
4.133
U4
V
+
0.400
U7
V
-2.533
R1
R4
R7
2Ω
2Ω
2Ω
V1
12 V
R2
2Ω
U2
+
3.733
V
-
R5
2Ω
U5
+
2.933
R6
R8
2Ω
2Ω
2Ω
-4.133
U3
V
+
-0.400
U6
V
V
-
R3
+
V
+
+
2.533
U8
V
V2
8V
The circuit with voltage measurements
according to corrected current directions
U1
+
4.133
U4
V
0.400
U7
V
V
+
2.533
R1
R4
R7
2Ω
2Ω
2Ω
V1
12 V
V
+
R2
2Ω
U2
+
3.733
V
-
R5
2Ω
U5
+
2.933
R3
R6
R8
2Ω
2Ω
2Ω
+
4.133
U3
V
+
0.400
U6
V
-
+
2.533
U8
V
V2
8V
The End