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# sovle this question.

Asked by yash 12th November 2020, 8:53 PM
Let us assume the current distribution in the circuit as shown in figure.

For the loop A-B-C-D-A , we write KVL as ,  4 i1 + 8 i3 = 12  V  .......................(1)

At node-B , by KCL , we have i3 = i1 + i2 .......................(2)

From eqn.(1) and eqn.(2) , we get ,  12 i1 + 8 i2 = 12   or 3 i1 + 2 i2 = 8  V  .......................(3)

For the loop B-E-F-C-B , we write KCL as ,  5 i2 + 8 i3 = 6  V  .......................(4)

From eqn.(6) and eqn.(2) , we get ,  8 i1 + 13 i2 = 6 V  .......................(5)

By solving eqn.(3)  and eqn.(5) , we get , i1 = ( 27/23 ) A   and  i2 = ( -6/23 ) A

Hence i3 = i1 + i2 = ( 21/23) A = 0.91 A
Answered by Expert 13th November 2020, 12:38 PM
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