1.

Two resistances wires, when joined in parallel, give an affective resistances of 6/5 Omega When one of the resistances wire is disconnected the effective resistances charges to 2 Omega. The resistance of the disconnected resistance wire is

Answer»

`4/5 OMEGA`
`2 Omega`
`6 Omega`
`3 Omega`

Solution :It is GIVEN that `R_1=2 Omega` and resistance of parallel grouping of TWO resistances `R_p=6/5 Omega`
As per parallel arrangement `1/R_p=1/R_1+1/R_2`
`implies 1/R_2=1/R_p-1/R_1=5/6-1/2=(5-3)/6=2/6=1/3 implies R_2=3 Omega`


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