Page 128 - Basic Electrical Engineering
P. 128

A mesh is a smallest loop in a network. KVL is applied to each mesh in terms
               of mesh currents instead of branch currents. As a convention, mesh currents
               are assumed to be flowing in the clockwise direction without branching out at

               the junctions. Applying KVL, the voltage equations are framed. By knowing
               the mesh currents, the branch currents can be determined. The procedure

               followed is explained through an example. Let us calculate the current
               flowing through the branches in the circuit given in Fig. 2.26.























                                                          Figure 2.26

               We have assumed loop currents I  and I  flowing in the clockwise direction
                                                       1
                                                               2
               as shown.
                  It may be noted that current flowing through the resistor R  is the algebraic
                                                                                         3
               sum of the two currents I  and I . Here I  is flowing in the downward
                                                                1
                                             1
                                                     2
               direction while I  is flowing in the upward direction.
                                   2
                  We will now write the voltage equations for the two loops applying KVL
               and then solve the equations. If the value of any mesh currents is calculated
               as negative, we will take the direction of that mesh current opposite to the

               assumed clockwise direction.
                  For loop DABCD, the voltage equation is


                               12 − 4 I  − 8(I  − I ) = 0
                                     1
                                                  2
                                             1
               or,                    3 I  − 2 I  = 3              (i)
                                     1
                                            2
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