Manora Caldera
Electrical Systems
Solutions to Additional Questions 3
1. Find the currents through each resistor and the voltage
(Ans: V0 = 33/5V)
+
in the circuit given in Figure 1.
+ -
-
-
+
+
+
-
Figure 1


Step 1
Currents are assigned to the two independent meshes in the clockwise direction as
I1 and I2
Step 2
Voltage polarities across each resistor are marked. (Note that the 2k
resistor is common to both meshes and hence will have two different voltage
polarities due to the two currents

Step 3: Apply KVL to each mesh
Considering the two mesh currents in the clockwise direction are I1 and I2. Applying
KVL to the meshes give
Mesh 1 (one on the left) :
2
4
2
3
0
Mesh 2 (one on the right) :
3 2
6 6
0
 Solving the two simultaneous equations will give
1
m A and I
10
I
and
in the downward direction.
11/10 mA
and
6
1
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2.
Using mesh analysis, determine the currents
and
in the circuit in Figure 2.
(Ans: 2mA, -6mA)
I4
I2
I3
Figure 2
Consider the four mesh currents in the clockwise direction as shown in the diagram.
12
Applying KVL to the meshes give
Mesh (one on the middle):
4
3
Mesh (one on the right):
12 2
0. This gives,
Mesh (one on the top):
6
3
Solving the equations give
12 10
As
3.
12
6
0. This gives, 9
2
3.33
0
,
Using mesh analysis, determine the current
I1
3
10
in the circuit in Figure 3.
(Ans: - 0.55mA)
I2
I3
I4
Figure 3
2
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Consider the four mesh currents in the clockwise direction as shown in the diagram.
2
Applying KVL to the meshes give
Mesh (one on the top left):
12 1
1
1
12
This gives, 3
Mesh (one on the bottom left):
1
1
(1)
0.
Mesh (one on the bottom right):
1
1
V
(2)
0.
(3)
2
Adding (2) and (3) gives,
4.
0
Also, using
4
It can be shown that
1.45
2
0
(4)
(5)
1.45
and hence
Using mesh analysis, determine the voltage
0.55
in the circuit in Figure 4.
(Ans: 14.4V)
I2
I1
I3
2
I4
Figure 4
Consider the four mesh currents in the clockwise direction as shown in the diagram.
6
Applying KVL to the meshes give
Mesh (one on the bottom left):
6 1
1
0
3
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This gives, 2
6
(1)
Mesh (one on the top right):
1
2
12 0.
As
6 , substituting this in (2) gives,
Mesh (one on the bottom right):
0
1
12 2
Using (1) and (3), it can be shown that
7.2
and hence
2
14.4
5.
(2)
2
(3)
Using mesh analysis, determine the voltage
in the circuit in Figure 5.
and the voltage across the current source
(Ans: V0 = 6V, 12.67V)
I1
I2
I4
I3
Figure 5
Consider the four mesh currents in the clockwise direction as shown in the diagram.
Applying KVL to the meshes give
Mesh (one on the bottom left):
2
6 1
0
6
This gives, 3
Mesh (one on the top left):
2
6 0.
Mesh (one on the top right):
1
12 0
(2) + (3) gives
2
6
Also,
4
(1)
(2)
(3)
(4)
(5)
4
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Using (4) + (5) gives,
6
Using (2),
and
2
6
2
12.67
Mesh (one on the bottom right):
1
12 1
0
This gives, 2
12
Using (1) and (6), it can be shown
1
As
(6)
0
6
and
6
6. Using nodal analysis, determine the voltage
in the circuit in Figure 6.
(Ans: V0 = -11V)
V1
V2
Figure 6
Applying KCL to Node 1 gives,
6
0
(1)
5
Rearranging (1) gives
3
36
(2)
Applying KCL to Node 2 gives,
4
0
Rearranging (3) gives
(1) + (2) gives,
4
(3)
3
8
44 giving
11
(2)
5
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7. Using nodal analysis, determine the current
in the circuit in Figure 7.
(Ans: I0 = 0.33mA)
Figure 7
As highlighted in the diagram, Node A has got four branches connected and Node B has
also got four branches connected.
Applying KCL to Node A gives,
1
2
0 This can be rearranged as 5
4
8
(1)
Applying KCL to Node B gives,
2
0 This can be rearranged as 3
Using (1) and (2) and solving for
gives,
8. Using nodal analysis, determine the voltage
2 , hence
6
12 (2)
0.33
in the circuit in Figure 8.
(Ans: V0 = 0.28V)
A
Figure 8
6
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Applying KCL to Node A gives,
6
0 This gives
12.82
.
As the current through the 3k resistor is
12.82
Hence,
9.
2.09
6
2.09
0.28
Using nodal analysis, determine the voltage
in the circuit in Figure 9.
(Ans: V0 = 7.2V)
I12V
I6v
1
2
3
Figure 9
Applying KCL to Node 1 gives,
0
(1)
Applying KCL to Node 2 gives,
6
0
(2)
Applying KCL to Node 3 gives,
0
(1
Also,
2
3
, 3
(3)
2
12
From (4) and (5), it can be shown
72
(5)
(4)
and
6
(6)
7.2
7
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10. Using nodal analysis, determine the voltage
in the circuit in Figure 10.
(Ans: V0 = -7.71V)
I12V
1
2
3
Figure 10
Applying KCL to Node 1 gives,
0
(1)
We do not need to apply KCL to Node 2 as we know the nodal voltage
6
Applying KCL to Node 3 gives,
0
(1
Also,
2
, 3
4
(2)
3
12
From (4) and (5), it can be shown
18
(3)
(4)
7.71
8