Solution 3.64
40 Ω
10 Ω
5 A
But at node A, io = i1 – i2 so that (1) becomes i1 = (16/6)i2
(2)
(3)
At node B, i3 + 0.2v0 = 2 + i1 (4)
0.2V
0
50 Ω
10 Ω
B
A
i1
i2
i3
i1
+ −
Solution 3.65
For mesh 1,
421 61212 III −−=
For mesh 2,
For mesh 3,
For mesh 4,
–I1 – I2 + 7I4 – 2I5 – 6 = 0 or
For mesh 5,
–I2 – I3 – 2I4 + 8I5 – 10 = 0 or
Casting (1) to (5) in matrix form gives
12
I
010612
−
Using MATLAB we input:
This leads to
Z =
12 -6 0 -1 0
V =
I =
2.1701
Solution 3.66
The mesh equations are obtained as follows.
or
–4I1 + 30I2 – 2I4 – 6I5 = –16 (2)
Putting (1) to (5) in matrix form
−
−−−
12
026430
Using MATLAB,
>> Z = [30,-4,-6,-2,0;
Z =
30 -4 -6 -2 0
V =
I =
-0.2779 A
Solution 3.67
Consider the circuit below.
+−
−
V35
025.035.0
Since we actually have four unknowns and only three equations, we need a constraint
equation.
5 A
−
−
5
325.335.0
Now we can use MATLAB to solve for V.
-0.2500 0.9500 -0.5000
I =
–5
V =
Let us now do a quick check at node 1.
–3(–30) + 0.1(–410.5) + 0.25(–410.5+194.74) + 5 =
Solution 3.68
Problem
Find the voltage Vo in the circuit of Fig. 3.112.
Solution
Consider the circuit below. There are two non-reference nodes.
3 A
3 A
Using MATLAB, we get,
Y =
0.1250 -0.1000
I =
7.0000
We can perform a simple check at node Vo,
Solution 3.69
For the circuit in Fig. 3.113, write the node voltage equations by inspection.
Figure 3.113
For Prob. 3.69.
Step 1. Assume that all conductance’s are in mS, all currents are in mA, and all voltages
are in volts.
Step 2. The node-voltage equations are:
−−
100
05.01.035.0
1
v
20 kΩ
v
1
v
2
v
3
Solution 3.70
With two equations and three unknowns, we need a constraint equation,
Solution 3.71
−−
30
549
We can now use MATLAB solve for our currents.
R =
V =
30
I =
Solution 3.72
By inspection, write the mesh-current equations for the circuit in Fig. 3.116.
Figure 3.116
For Prob. 3.72.
Step 1. First we write the resistance equations by inspection.
R11 = 20 + 10 = 30, R22 = 20 + 30 = 50, R33 = 10 + 20 = 30, R44 = 20 + 30 = 50,
Step 2. Hence the mesh–current equations are:
−
40
010030
i
20
Ω
Solution 3.73
Write the mesh-current equations for the circuit in Fig. 3.117.
10 Ω
+ −
10 Ω
10 Ω
Figure 3.117
For Prob. 3.73.
Solution
Loop 1. –15 + 10i1 + 10(i1–i2) + 10(i1–i3) = 0 or 30i1 – 10i2 – 10i3 = 15
10 Ω
+ −
10 Ω
10 Ω
1
30 10 10 0 15
i
−−
Solution 3.74
R11 = R1 + R4 + R6, R22 = R2 + R4 + R5, R33 = R6 + R7 + R8,
−
2
1
V
V
Solution 3.75
* Schematics Netlist *
R_R4 $N_0002 $N_0001 30
i3
Solution 3.76
* Schematics Netlist *
I_I2 0 $N_0001 DC 4A
R_R1 $N_0002 $N_0001 0.25
Solution 3.77
As a check we can write the nodal equations,