ICSE Class 10 Answered

Asked by keshavendu35 | 26 Sep, 2018, 10:25: AM

Loop BDCB :- 3×i1 + 4×(i1+i3) = 7×i1 + 4×i3 = 6 .....................(1)
Loop BACB:- 3×i2 - 3×i3 = 6 ....................................................(2)
Loop BADB -6×i3 + 3×i1 = 0 or i1 = 2×i3 .................................(3)
solving the above eqns., we get i1 = 2/3 A , i2 = 7/3 A , i3 = 1/3 A
Current drawn from battery, i = i1 + i2 = (2/3) + (7/3) = 3 A
Answered by Thiyagarajan K | 26 Sep, 2018, 11:56: AM
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