I =
2.0000
Solution 10.17
Consider the circuit below.
At node 1,
20100
2111
VVVV −
−°∠
(1)
At node 2,
(2)
From (1) and (2),
+
°∠
1
)j3(5.05.0–
20100
V
°∠=
∆
=08.13–74.64
1
V
Solution 10.18
Consider the circuit shown below.
At node 1,
454
211
−
+=°∠ VVV
At node 2,
22
21
−VV
VV
(2)
Substituting (2) into (1),
22 )3j4(
3j104
)41j12(
)3j29(45200 VV −−
−
+
−=°∠
j6 Ω
4 Ω
8 Ω
V
1
V
2
j5 Ω
Solution 10.19
We have a supernode as shown in the circuit below.
At the supernode,
31
12
23 VV
VV
VV −
−
At node 3,
VVVV
−
Subtracting (2) from (1),
But at the supernode,
Substituting (4) into (3),
)12(j2.10
11
−+= VV
j2 Ω
Solution 10.20
The circuit is converted to its frequency-domain equivalent circuit as shown below.
Let
Lj
C
L
1
||Lj
ω
=
=
ω=Z
R
Solution 10.21
1
Cj
1
o
ω
V
(b)
RCjLC1
LC
1
Lj
2
2
i
o
ω+ω−
ω−
=
ω
=
V
V
Solution 10.22
Consider the circuit in the frequency domain as shown below.
Let
Cj
1
||)LjR( 2ω
ω+=Z
CRjLC1
LjR
2
2
ω+ω−
ω+
Z
V
R1
Solution 10.23
0CVj
1
V
VV s=ω+
+
−
Solution 10.24
Design a problem to help other students to better understand mesh analysis.
Problem
Solution
Consider the circuit as shown below.
2 Ω
+
Solution 10.25
2=ω
°∠→ 010)t2cos(10
The circuit is shown below.
For loop 1,
(1)
For loop 2,
(2)
In matrix form (1) and (2) become
j4 Ω
4 Ω
Solution 10.26
ω
→ = =
3
0.4 10 0.4 400H jL j x j
The circuit becomes that shown below.
2 kΩ
–j1000
Io
For loop 1,
(2)
In matrix form, (1) and (2) become
Solution 10.27
Using mesh analysis, find I1 and I2 in the circuit of Fig. 10.75 as shown in the text.
Figure 10.75
For Prob. 10.27.
Solution
For mesh 1,
For mesh 2,
From (1) and (2),
°∠=+=∆ 56.116472.4j42–
Solution 10.28
The frequency-domain version of the circuit is shown below, where
1 j4 j4 1
-j0.25
Applying mesh analysis,
From (1) and (2), we obtain
Solving this leads to
o
2
o
192114.4I,07.41741.2I ∠=−∠=
Hence,
Solution 10.29
Using Fig. 10.77, design a problem to help other students better understand mesh
analysis.
Problem
By using mesh analysis, find I1 and I2 in the circuit depicted in Fig. 10.77.
Figure 10.77
(1)
For mesh 2,
(2)
From (1) and (2),
++
°∠
j)(2–5j5
2030
I
°∠=°∠°∠=∆ 56.4608.67)56.26356.2)(2030(
2
Solution 10.30
ω
−
→ = =
3
300 100 300 10 30mH j L j x x j
ω
−
→ = =
3
200 100 200 10 20mH j L j x x j
ω
−
→ = =
3
400 100 400 10 40mH j L j x x j
The circuit becomes that shown below.
j40 j20
20 Ω
For mesh 1,
We put (1) to (3) in matrix form.
−+
12j
I
03j3j2
1
This is an excellent candidate for MATLAB.
Z =
>> V=[12i;0;-8]
V =
I =
2.0557 + 3.5651i
Solution 10.31
Use mesh analysis to determine current Io in the circuit of Fig. 10.79 below.
Figure 10.79
For Prob. 10.31.
Solution
Consider the network shown below.
For loop 1,
040)4080(12050–
=+−+°∠ II jj
For loop 2,
For loop 3,
32
From (2),
Io
j60 Ω
20 Ω
80 Ω
From (1) and (4),
−
°∠
1
4)2(4
1205
I
jj
−
4j–4j8