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Question-90632




Question Number 90632 by awlia last updated on 25/Apr/20
Answered by ajfour last updated on 25/Apr/20
−i(10Ω)+10V−2i(5Ω)+5V=0  (for the left loop or right loop)  ⇒  3i=((15V)/(15Ω)) = 1A  current in 5Ω is = 2i = (2/3)A.
i(10Ω)+10V2i(5Ω)+5V=0(fortheleftlooporrightloop)3i=15V15Ω=1Acurrentin5Ωis=2i=23A.

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