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Determine the current in each branch of the network shown in Fig. |
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Answer» SOLUTION :As per Kirchhoff.s first rule the distribution of current in various branches is shown in Fig. Now applying Kirchhoff.s SECOND law to mesh ABDA, we have ` - 10.I_1 - 5.I_2 + 5.(I - I_1) = 0` or` + 5I - 15I_1 - 5I_2 = `...(i) Again for mesh BDCB , we have ` - 5.I_2 - 10.(I - I_1 + I_2) + 5.(I_1 - I_2) = 0` or`-10 I + 15I_1 - 20I_2 = 0`...(ii) and for mesh ABCEA , we have ` - 10 . I_1 - 5.(I_1 - I_2) - 10.I + 10 = 0` or`10I + 15I_1 - 5I_2 = 10`....(iii) Solving these three equations , we FIND that `I =10/17 , I_1 = 4/17 A and I_2 = -2/17A` Hence , electric current in branch `AB , I_1 = 4/17`A, in branch `AD, (I - I_1) = 6/17 A` , in branch BD , `I_2 = -2/17A` , in branch BC , `(I_1 -I_2) = 6/17 A,` in branch `DC,(I - I_1 + I_2) = 4/17A` and the total current drawn from battery `I = 10/17 A` |
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