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Definition:

Consider a connected graph G with b branches and n

_{t}nodes. Select any arbitrary tree. The tree will contain n = n_{t}- 1 tree branches (twigs) and l = (b - n) link branches. Every link defines a fundamental loop of the network. Let us take the example of the graph shown in Fig. a.**Fig.**

Let T be a tree of G as shown in Fig. b. The number of fundamental loops of this graph will be (b - n) = 3. the three f-loop l

_{1}, l_{2}and l_{3}are shown in Fig. c. In order to apply KVL to each fundamental loop, we take reference direction of the loop which coincides with the reference direction of the link defining the loop.
l

_{1}: v_{1}+ v_{2}+ v_{5}+ = 0
l

_{2}: v_{2}+ v_{3}+ v_{4}= 0
l

_{3 : }v_{1 + }v_{3 + }v_{6}= 0
In matrix form, we can write

B

_{f}v_{b}= 0 (KVL)
Where is an matrix called the fundamental loop matrix or tie-set matrix

B

_{f}= [b_{kj}]
Where b

_{kj}are the elements of b_{f}the (k,j) element of the matrix is defined as follows:
b

_{kj }= 1 when branch b_{j}is in the f-loop l_{k}and their reference directions (orientations) coincide
b

_{kj}= -1 when branch b_{j}is in the f-loop l_{k}and has opposite orientation
b

_{kj}= 0 when branch is not in the f-loopExample:

In the graph shown in Fig. a tree consisting of branches 4, 5, 6, 7, 8 is chosen, as shown by heavy lines. Write the fundamental loop matrix of the graph.

**SOLUTION**

The fundamental loops defined by links {1, 2, 3} and their orientations are shown in Figs, c and d. Consider loop l

b

_{1}. It contains branches {1, 6, 8, 5}. The orientation of loop l_{1}is given the same orientation as its defining link 1. Therefore the element b_{11}is written 1. The directions of branches 6, 8 and 5 in l_{1}are the same as l_{1}. therefore the entries**FIG.**b

_{16},b_{18}and b_{15 }are each equal to 1. Since branches 2, 3, 4 and 7 are not in loop b_{13}= 0,b_{14}= 0,b_{17}= 0.
Loop l

_{2}contains branches {2, 6, 7}. The orientation of l_{2}is given the same orientation as its defining link. 2. Therefore the element b_{22}= 1. The orientations of branches 6 and 7 are opposite to the orientation of l_{2}consequently, b_{126}= -1 and b_{27}= -1 Since branches {1, 3, 4, 5, 8} are not contained in l_{2}, b_{21}= 0,b_{23}= 0,b_{24}= 0,b_{25}= 0,b_{28}= 0.
Loop l

_{3 }has branches (3, 4, 8, 7). The orientation of l_{3}is the same as that of its defining link 3.Therefore the element b_{33}= 1 The directions of branches 4 and 7 coincide with the orientation of l_{3}Hence b_{34}= 1, b_{37}= 1. The orientation of branch 8 does not coincide with the orientation of l_{3}. Hence b_{38}= -1. Since branches {1, 2, 5, 6} are not contained in l_{3},b_{31}= 0, b_{32}= 0, b_{35}= 0, b_{36}= 0. Thus, we obtain the following matrix:
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