Network Analysis IV
Mesh Equations – Two Loops
Circuit overview
R1
A
C
10Ω
R2
E
15Ω
_
+
V1
40 V
_
Mesh A
IA
R3
10Ω
V2
20 V
Mesh B
IB
+
D
B
F
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 m.
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. Which component has opposing
mesh currents (if any)?
Listing of steps (cont.)
d. Write the mesh equation for mesh
A.
e. Write the mesh equation for mesh
B.
f. Solve for the currents IA and IB using
any of the methods for the solution
of simultaneous equations.
Listing of steps (cont.)
g. Determine the values of currents I1, I2 and
I3.
h. Are the assumed directions of mesh A and
mesh B currents correct? How do you
know?
i. What is the direction of current I3 through
R3?
j. Solve for the voltage drops VR1, VR2 and VR3.
Listing of steps (cont.)
k. Using the final solutions for VR1, VR2 and VR3,
write a KVL equation for the loop ACDBA
going clockwise from point A.
l. Using the final solutions for VR1, VR2 and VR3,
write a KVL equation for the loop EFDCE
going clockwise from point E.
m. Using the final solutions (and directions) for
I1, I2 and I3, write a KCL equation for the
currents at point C.
Step a solution
R1
A
C
10Ω
+
V1
40 V
_
Mesh A
IA
R3
10Ω
D
B
a. Identify the
components through
which the mesh
current IA flows.
• IA flows through V1, R1
and R3 (not necessarily
in that order).
Step b solution
C
R2
b. Identify the
components through
which the mesh
current IB flows.
E
15Ω
_
R3
10Ω
V2
20 V
Mesh B
IB
+
D
F
• IB flows through V2, R2
and R3 (not necessarily
in that order).
Step c solution
R1
A
C
10Ω
R2
E
15Ω
_
+
V1
40 V
_
Mesh A
IA
R3
10Ω
V2
20 V
Mesh B
IB
+
D
B
c. Which component has
opposing mesh
currents (if any)?
F
• The component with
the opposing mesh
currents is R3 (down
through R3 from IA and
up through R3 from IB).
Step d solution
R1
A
C
10Ω
+
V1
40 V
_
Mesh A
IA
R3
10Ω
D
B
d. Write the mesh
equation for mesh A.
• 20IA – 10IB = -40V
Step e solution
C
R2
e. Write the mesh
equation for mesh B.
E
15Ω
_
R3
10Ω
V2
20 V
Mesh B
IB
+
D
F
• -10IA + 25IB = -20V
Step f solution
Original equations:
• Step d:
20IA – 10IB = -40V
• Step e:
-10IA + 25IB = -20V
f.
Solve for the currents IA
and IB using any of the
methods for the solution
of simultaneous
equations.
• Step 1  Divide equation
from step d by 2 (i.e.
make values of IA the
same)
10IA – 5IB = -20V
Step f solution (cont.)
• Step 2  Add equations
10IA – 5IB = -20V
-10IA + 25IB = -20V
20IB = -40V
Step f solution (cont.)
• Step 3  Solve for IB by dividing multiplier
𝟐𝟎𝑰𝑩
𝟐𝟎
=
−𝟒𝟎
𝟐𝟎
∴ 𝑰𝑩 = −𝟐𝐀
Step f solution (cont.)
• Step 4  Substitute IB into either equation to
solve for IA (we will use one from step d)
20IA – 10(-2A) = -40  20IA + 20 = -40
Step f solution (cont.)
• Step 5  Move all the known to the right
side, which will leave our unknown on the left
20IA = -60
• Step 6  Divide multiplier to solve for IA
𝟐𝟎𝑰𝑨
𝟐𝟎
=
−𝟔𝟎
𝟐𝟎
∴ 𝑰𝑨 = −𝟑𝐀
Step g solution
g. Determine the values of currents I1, I2 and I3.
• I1 = IA = -3A
• I2 = IB = -2A
• I3 = IB – IA = -2A – (-3A) = -2A + 3A = 1A
Step h solution
h. Are the assumed directions of mesh A and
mesh B currents correct? How do you know?
• The assumed currents are not correct as they
are negative in value. The more appropriate
statement would be a reversal of the current
directions.
Step i solution
R1
A
C
10Ω
R2
i. What is the direction
of current I3 through
R3?
E
15Ω
_
+
V1
40 V
_
R3
10Ω
V2
20 V
+
D
B
F
• The overall direction of
I3 is up through R3
(counterclockwise for
mesh A and clockwise
for mesh B)
Step j solution
• Solve for the voltage drops VR1, VR2 and VR3.
• VR1 = I1R1 = 3A(10Ω) = 30V
• VR2 = I2R2 = 2A(15Ω) = 30V
• VR3 = I3R3 = 1A(10Ω) = 10V
Step k solution
30V
_
+ R1
A
C
10Ω
+
V1
40 V
_
10V
+
R3
10Ω
_
D
B
k. Using the final
solutions for VR1, VR2
and VR3, write a KVL
equation for the loop
ACDBA going clockwise
from point A.
30V + 10V – 40V = 0
Step l solution
30V
C
+ R2
_
E
15Ω
+
10V
_
_
R3
10Ω
V2
20 V
+
D
F
l. Using the final
solutions for VR1, VR2
and VR3, write a KVL
equation for the loop
EFDCE going clockwise
from point E.
-20V – 10V + 30V = 0
Step m solution
• Using the final solutions (and directions) for I1,
I2 and I3, write a KCL equation for the currents
at point C.
• I3 + I2 = I1; this translates into the KCL equation
 I3 + I2 – I1 = 0 ∴ 1A + 2A – 3A = 0
(I3 and I2 flow into node C, while I1 flows out of
node C).
The circuit with directions and voltage
measurements
+
30.000
V
+
30.000
R1
A
C
10Ω
R2
V
E
15Ω
_
+
V1
40 V
_
+
10.000
V
R3
10Ω
V2
20 V
-
+
D
B
F
The End