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B
4A
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C
2A
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D
3A
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Solution
The correct option is C2A −i1+0×ixy+3i2=0,i.e.,i1=3i2...(i) Also −2(i1−ixy)+4(i2+ixy)=0 ⇒2i1−4i2=6ixy...(ii) Also VAB=1×i1+2(i1−ixy)=0⇒50=i1+2(i1−ixy) =3i1−2ixy...(iii) Solving (i), (ii) and (iii), ixy=2A