1.

63. For the network shown in the figure thethe current i is :-2Ω4Ω4Ω3Ω6Ω18V5V

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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