Page 130 - Basic Electrical Engineering
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Applying KVL in mesh II,
− 31 I − 2 − 5I + 6 + 6(I − I ) = 0
2
2
2
1
or, − 6I + 14I = 4 (ii)
1
2
Adding eqs. (i) and (ii)
11 I = 6
2
or,
and
Current through the 6 Ω resistor is (I − I ) which is equal to .
1
2
Example 2.9 A network with three meshes has been shown in Fig. 2.28.
Applying Maxwell’s mesh current method determine the value of the
unknown voltage, V for which the mesh current, I will be zero.
1
Figure 2.28
