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Three resistor 4Omega , 6Omega and 8 Ome...

Three resistor `4Omega , 6Omega and 8 Omega`, are combined in parallel. What is the total resistance of the combination ?

Text Solution

Verified by Experts

Given `R_(1)=4Omega,R_(2)=6Omega,R_(3)=8Omega,`
WKT, for parallel combination,
`(1)/(R_(8))=(1)/(R_(1))+(1)/(R_(2))+(1)/(R_(3))`
i.e, `(1)/(R_(8))=(1)/(4)+(1)/(6)+(1)/(8)=(6+4+3)/(24)=(13)/(24)`
Hence equivalent resistance in parallel combination is `R_(p)=(24)/(13)=1.84Omega`
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