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The rate of energy consumed in 5Omega re...

The rate of energy consumed in `5Omega` resistance [shown in Fig . 3 .3] is `10 J.s^(-1)` .What will be the rate of energy consumed in `4Omega` resistance ?

Text Solution

Verified by Experts

In the adjoining figure if , `V_(A)-V_(B)=V, "then" `
`I_(1)=(V)/(5)andI_(2)=(V)/(4+6)=(V)/(10)`

`therefore("Energy consumed in 1s in" 4Omega"resistance")/("Energy consumed in 1s in" 5Omega"resistance") `
`=(I_(2)^(2).4)/(I_(1)^(2).5)=(((V)/(10))^(2).4)/(((V)/(5))^(2).5)=(1)/(5)`
So , energy cosumed in 1 s in `4Omega` resistance
`=(1)/(5)xx"energy consumed in" 5Omega`resistance
`=(1)/(5)xx10=2J` i.e , the rate of energy comsumed in `4Omega` resistance is `2J.s^(-1)`
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