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How can three resistor of resistances 2 Omega, 3 Omegaand 6 Omegabe connected to give a total resistance of 1 Omega ?

Answer»

Solution : In order to OBTAIN a total resistance of `1Omega` from three resistor of `2Omega, 3 Omega and 6 Omega`, all the three resistors should be comnected inparallel.This is because in a parallel COMBINATION,
`1/R_p = 1/R_1 +1/R_2 + 1/R_3`
`=1/2 + 1/3 +1/6`
`= (3 + 2 +1)/6`
`= 6/6`
`=1`
`:. R_p =1 Omega`
Hence, the ARRANGEMENT of three resistors `2Omega, 3Omega and 4 Omega` which gives total resistance `1 Omega` can be represented as follows :


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