1.
(c)
(d)
(e)
(f)
Answer the following questions about the circuit below.
(a) How many branches are there?
___________
(b) How many nodes are there? ___________
(g)
How many meshes are there?
How many supernodes are there?
___________
___________
10
6
2
5
How many supermeshes are there? ___________ 1
In writing nodal equations for this circuit, how many equations should you write for the sum of currents leaving a node or supernode equals zero (KCL)? ___________ 2
In writing mesh equations for this circuit, how many equations should you write for the sum of voltages around a mesh or supermesh equals zero (KVL)? ___________ 4
30
Ω
20 Ω
5
Ω
-12V
4v
1
40
Ω
10V 20 Ω v
1
2A 15 Ω
2.
Fill in the blanks below in the nodal and mesh equations.
3
+ v
2 = 0
5
+
12
+ v
2
12
+ v
3
5
=
= 0
6
+
4
+ v
2
5
+ v
5 = 0
5 +
( i
1 i
2
)
+ 4 + 6 + i
1
= 0 i
2
( i
4
+ + 4 + 12 = 0 i
3
)
20 + 10 = 0
5
Ω
v
5 v
2
5
Ω v
1
12
Ω
i
2
v
3
i
1
3
Ω
4
Ω
2A i
3
20V v
4
6
Ω
i
4
10
Ω

3.
v
1
3 v
2
+ v v
1
1 v
4
6
5 v
3
12 v
2
+ v
4
5
+
+
4 v
2 v v v
2
3
3
=
12
4
+
0 v
3 v v
4
5
+
=
5 v
2 v
2
2
5 v
5
+ v
5
10
= 0
= 0
5 i
1
5 i
2
6
( i
4
+
+
12
20
( i
+
1 i
3
)
4
( i
2 i
20
2
)
+
+ 4 i
3
10
(
) i
4 i
+
3
12
= 0 i
2
(
) i
2
+ 6 i
1
( i
)
3
= 0 i
4
)
+ 3 i
1
= 0
Find the numerical value of R.
R = ________________
(The numerical value of the voltage in volts across the dependent voltage source is equal to the negative of half of the numerical value of the current i in amperes.)
4 Ω
R i
12V 8 Ω −i/2
0.5A
Assign
v
1 node.
as the node voltage at the intersection of the three resistors and the bottom node as the reference v
1
4
12
+ v
1
8
+ v
1
( i / 2
)
R
= 0 i =
12 v
1
4
0.5
= v
1
( i / 2
)
R v
1
4
12
+ v
1
8
+ 0.5
= 0
( ) v
1
= 2.5
v
1
= 20 / 3 V i =
12 20 / 3
4
= 4 / 3 A , 0.5
=
20 / 3
(
2 / 3
)
R
R =
20 / 3
(
2 / 3
)
0.5
=
(
44 / 3
)
Voltage drop across 4 ohm resistor is 16/3 V. Current through the 8 ohm resistor is 5/6 A. Voltage drop across R is 22/3 V. Dependent voltage source voltage is –2/3 V. KVL and KCL are satisfied everywhere.
Alternate Solution:
Assign two mesh currents,
i
2
i
1
in the left mesh and
= 0.5
. The two mesh equations are
i
2
8
(
12 + 4 i
1 i
2 i
1
)
+ 8 i
1
+ Ri
2
(
in the right mesh. Then immediately we know that i
2
)
= 0
+
( i / 2
)
= 0
From the first mesh equation
12 + 4 i
1
+ 8
( i
1
0.5
)
= 0 12 i
1
= 16 i
1
= 4 / 3 i = 4 / 3
From the second mesh equation
(
4 / 3
)
+ R 1 / 2 +
(
2 / 3
)
= 0 R = 2 22 / 3
)
=
(
44 / 3
)