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When two resistance are connected in parallel then the equivalent resistance is `(6)/(5)Omega` When one of the resistance is removed then the effective resistance is `2Omega`. The resistance of the wire removed will beA. 3 ohmB. 2 ohmC. `(3)/(5)ohm`D. `(6)/(5)ohm`

Answer» Correct Answer - A
`(R_(1)R_(2))/(R_(1)+R_(2))=(6)/(5)`. If `R_(2)` is removed `R_(1)=2Omega`
`(2R_(2))/(2+R_(2))=(6)/(5)implies5R_(2)=6+3R_(2)impliesR_(2)=3Omega`


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