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63. For the network shown in the figure thethe current i is :-2Ω4Ω4Ω3Ω6Ω18V5V |
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Answer» This arrangement is a wheat stone bridge as 4/6 = 2/3 .. so. here , the middle resistance has no. current. so, the 4Ω and 2Ω , will combine in series add up to 6Ω and the lower 6 and 3 will add up to 9Ω now this 6 and 9 are in parallel to each other so, Req = 6*9/(9+6) = 54/15 Ω so I = V/R = 15V/54 = 5V/18 . |
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