Network Analysis and Synthesis, 1st Edition by Pearson

Network Theorems and Applications 215

=

∠°

−

=

∠°

∠−

=

109 992 36 89

5 9199 656

109 992 36 86

8 836 47 93

12

..

..

..

..

.

j

444 84 79∠°.A

Example 5.25 Verify the reciprocity theorem for

the network shown in Figure 5.98.

Solution: Let us find current in branch CD, that is,

I due to the voltage source in branch AB as in

Figure 5.99.

Applying KVL in mesh I, we get the following:

3 (i

1

− i

2

) = 10

or

ii

12

– =

10

3

i

1

− i

2

= 3.33 (5.26)

Applying KVL in mesh II, the following form can

be obtained:

3(i

2

− i

1

) + (−j10) (i

2

− i

3

) = 0

or − 3i

1

+ (3 − j10) i

2

+ j10 i

3

= 0 (5.27)

Applying KVL in mesh III, the equation can be written as follows:

− J10 (i

3

− i

2

) + (2 + j5) i

3