Current P through a certain non-metallic rod is given as `i_(0)=0.2V^((3)/(2))` where V is potential difference across it. The rod is connected in series with a resistance to a 12 volt ideal bettery. What should be resistance ( in `Omega`) so that power dissipated in the resistance is twice that of rod.

Answer

5

Explanation

`p_(res)=2P_(rod)`
`(12-v)i=2Vi`
`V=4` volt
`I^(2)R=2vi`
`(0.2)^(2)V^(3)R=2V(0.2)V^((3)/(2))`
`R=Omega`