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

Three resistor 4 `Omega`, 6 `Omega` and 8 `Omega`, are combined in parallel.If the combination is connected to a battery of emf 25 V and negligible Internal resistance, then determine the current through each resistor and total current drawn from the battery.

Text Solution

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Given `R_(1)=4W,R_(2)=6W,R_(3)=8W,`
Main current
`I=(E)/(R_(eq)+r)\\r=0`
i.e, `I=(25)/(1.846)`
i.e, `I=13.543A`
Total current drawn from the battery is 13.543 A.
Branch current `I_(1)=(25)/(4)=6.250A`
`I_(2)=(25)/(6)=4.168A`
`&" "I_(3)=(25)/(8)=3.125A`
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