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Asked by yash | 12 Nov, 2020, 20:53: PM
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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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