PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

Direct Current and Transient Analysis 157

>> disp(‘The Symbolic current solutions are given by:’)

>> [I1 I2 I3] = solve(‘15*I1-5*I2=10’,’-5*I1+35*I2=0’,’I1-I2=I3’)

>> disp(‘*********************************’)

************************************************

The Symbolic current solutions are given by:

I1 = 7/10

I2 = 1/10

I3 = 3/5

************************************************

Note that the results obtained for parts i and ii are equivalent to the solutions obtained in

Example 2.13. Also note that the current I 3 in Example 2.13 is labeled I 2 in Example 2.14.

Example 2.15

Show that the circuit diagrams in Figure 2.54 are equivalent, as seen through the termi-

nals labeled a and b.

ANALYTICAL Solution

The circuit diagrams in Figures 2.54a and 2.54b are equivalent due to the equivalent

substitutions shown in Figure 2.55.

The circuit diagram shown in Figure 2.54b is equivalent to the circuit diagram shown

in Figure 2.54c due to the equivalent substitutions indicated in Figure 2.56.

The circuit diagram in Figure 2.54c is equivalent to the diagram shown in Figure

2.54d due to the equivalencies shown in Figure 2.57.

Finally, the diagram in Figure 2.54d is equivalent to the diagram in Figure 2.54e due

to the substitution indicated in Figure 2.58.

The equivalencies shown can be evaluated by using the MATLAB programs of

Examples 2.2 and 2.3 and are left as an exercise for the reader to verify.

FIGURE 2.54

Diagrams of Example 2.15.

a b

(c)

1

3

12

3

a^5 b

(e)

a^14 b

(d)

a^1 b

75

6

6

3

(b)

a b

(a)

15

10

30

7

12

4

2

6

4

1