In a part of circuit shown in the figure, the rate of heat dissipation in `4 Oemga` resistor is `100 Js^(-1)`. Calculate the heat dissipated in the `3 Omega` resistor in 10 second.

Explanation

Let `I_(1)` be the current in series combination of `R_(1)` and `R_(2)` and `I_(2)` be the current through `R_(3)`.
pot. Diff. across `(R_(1) + R_(2))` = pot. Diff. across `R_(3)`
`(4 + 2 ) I_(1)= 3 I_(2)` or `I_(2) = 2 I_(1)`
Rate of heat dissipation in `4 Omega` resistior
`I_(1)^(2) R_(1) = I_(1)^(2) xx 4 = 100 Js^(-1)` (Given)
So `I_(1) = sqrt((100)/(4)) = 5A`
and `I_(2) = 2 I_(1)=2 xx 5 = 10A`
Heat dissipated in `3 Omega` resistor in 10 s
`I_(2)^(2) R_(2) t= (10)^(2) xx 3 xx 10 = 3000 J` .