Solution 3.36
Use mesh analysis to obtain ia, ib, and ic in the circuit shown in Fig. 3.84.
10 Ω
15 Ω
20 Ω
ib
Figure 3.84
For Prob. 3.36.
Step 1. Establish two unknown loop currents and write the mesh equations. Then solve
the mesh equations for the two unknown loop currents which will allow us to
solve for the unknown branch currents.
15 Ω
5 Ω
10 Ω
20 Ω
ia
ib
ic
Step 2. The matrix equation is,
30 10

Solution 3.37
20Ω
4 Ω
Applying mesh analysis to loops 1 and 2, we get,
20Ω
6 Ω
Solution 3.38
Consider the circuit below with the mesh currents.
1 Ω
1 Ω
1 Ω
4 Ω
I1 = –5 A (1)
But, we need one more equation, so we use the constraint equation –I3 + I4 = 10. This
now gives us three equations with three unknowns.
−
−
5.27
I
107
2
4 Ω
3 Ω
Io
2 Ω
2 Ω
Z =
V =
–27.5
I =
Check using the super mesh (equation (3)):
Solution 3.39
Solution
Given R1 = 4 kΩ, R2 = 2 kΩ, and R3 = 2 kΩ, determine the value of Ix using mesh
analysis.
R1
R2
Ix
Solution 3.40
4 kΩ
4 kΩ
Assume all currents are in mA and apply mesh analysis for mesh 1.
for mesh 2,
for mesh 3,
6 kΩ
2 kΩ
6 kΩ
i
2
Solution 3.41
5 Ω
1 Ω
4 Ω
2 Ω
For loop 3,
We put (1), (2), and (3) in matrix form,
−
3
i
016
1
i3
10 Ω
i2
i
0
316
−
Solution 3.42
Problem
Determine the mesh currents in the circuit of Fig. 3.88.
Figure 3.88
Solution
For mesh 1,
For mesh 3,
Putting eqs. (1) to (3) in matrix form, we get
I
−
12
03050
Using Matlab,
48.0
Solution 3.43
80 V
20 Ω
30 Ω
30 Ω
For loop 1,
For loop 2,
a
20 Ω
20 Ω
b
Solution 3.44
Use mesh analysis to obtain io in the circuit of Fig. 3.90.
Figure 3.90
For Prob. 3.44.
Step 1. We need to redraw the circuit using a supermesh. Next we identify our unknown
loop currents. Then we write our mesh equations and write the equation
incorporating the current from the current source.
50 V
50 V
Step 2. Since we need i2 and i3 let us use the constraint equation, i1 = i2–4, to allow us to
solve for i2 and i3.
Using the first two equations we get,
200 100

Solution 3.45
1 Ω
i1
i2
3 Ω
For loop 2, 10i2 – 3i1 – 6i4 = 0 (2)
i3
i4
6 Ω
2 Ω
4 Ω
8 Ω
Solution 3.46
Calculate the mesh currents i1 and i2 in Fig. 3.92.
Figure 3.92
For Prob. 3.46.
Step 1. Loop 1 –40 + 10i1 + 10(i1–i2) = 0 and for loop 2 10(i2–i1) + 10 i2 + 2vo =0.
10 Ω
10 Ω
+ vo
−
Solution 3.47
First, transform the current sources as shown below.
– 6V +
2
Ω
Ω
Ω
Ω
Ω
Ω
– –
For mesh 1,
For mesh 2,
321123
Putting (1) to (3) in matrix form, we obtain
I
−−
10
417
1
Using MATLAB,
2
But
=−=→
−
=111 420
4
20 IV
V
I
10 V
Solution 3.48
We apply mesh analysis and let the mesh currents be in mA.
3k
Ω
I4
4k
Ω
2k
Ω
5k
Ω
Ω
Ω
–
For mesh 1,
For mesh 2,
For mesh 3,
For mesh 4,
Putting (1) to (4) in matrix form gives
2
I
4015
1
−−
Using MATLAB,
608.3
Solution 3.49
Find vo and io in the circuit of Fig. 3.94.
Figure 3.94
For Prob. 3.49.
Step 1. First we note that we have three unknown loop currents but we can only write two
mesh equations (one is a supermesh). So we will need a constraint equation or io
and three unknowns.
2 Ω
2i
3 Ω
i
1
0
i
2
(a)
Step 2. 3i1 + 2i2 – 3i3 = –54 = (3–2)i1 – 3i3 or i1 – 3i3 = –54.