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Three resistors 2Omega, 4Omega and 5Omeg...

Three resistors `2Omega, 4Omega` and `5Omega` are connected in parallel. What is the total resistance of the combination?

Text Solution

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Given `R_(1) = 2Omega, R_(2) = 4Omega, R_(3) = 5Omega`
For a parallel combination, `1/(R_(p))=1/(R_(1))+1/(R_(2))+1/(R_(3))`
i.e., `1/(R_(P))=1/2+1/4+1/5=(10+5+4)/20=19/20`
`thereforeR_(P)=20/19Omega=1.0052 Omega`
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SUBHASH PUBLICATION-CURRENT ELECTRICITY-NUMERICAL PROBLEMS AND ANSWERS
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