JEE Class main Answered
sovle this question.
Asked by yash | 12 Nov, 2020, 08:53: PM
Expert Answer
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 Thiyagarajan K | 13 Nov, 2020, 12:38: PM
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