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NOTE These equations can be greatly simplified if we use the method of
elimination of variables. This is essentially the method of nodes analysis cov-
ered in circuit theory courses. At this time, our purpose is to show a direct
numerical method for obtaining the solutions.
If we form column vector VI, the top three components referring to the
voltages V , V , V , and the bottom three components referring to the cur-
3
2
1
rents I , I , I , then the following script M-file provides the solution to the
2
1
3
above circuit:
M=[1 0 0 0 0 0;1 -1 0 -50 0 0;0 1 -1 0 -100 0;...
0 1 0 0 0 -300;0 0 1 0 -2 0;0 0 0 1 -1 -1];
Vs=[5;0;0;0;0;0];
VI=M\Vs

In-Class Exercise

Pb. 8.4 Use the same technique as shown in Example 8.6 to solve for the
potentials and currents in the circuit given in Figure 8.2.

10 V
V 1 100 Ω V 2 50 Ω V 3 V 4
I 3
I 1 I 2
I 4 I 5
7 V 200 Ω 100 Ω 300 kΩ

FIGURE 8.2
Circuit of Pb. 8.4.

8.7.2 dc Circuit Design
In design problems, we are usually faced with the reverse problem of the
direct analysis problem, such as the one solved in Section 8.7.1.

Example 8.7
Find the value of the lamp resistor in Figure 8.1, so that the current flowing
through it is given, a priori.

Solution: We approach this problem by defining a function file for the rele-
vant current. In this case, it is

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