The current `I` through a rod of a certain metallic oxide is given by `I=0.2 V^(5//2)`, where `V` is the potential difference across it. The rod is connected in series with a resistance to a `6V` battery of negligible internal resistance. What value should the series resistance have so that:
(i) the current in the circuit is `0.44`
(ii) the power dissipated in the rod is twice that dissipated in the resistance.

Answer

`(i) 10.52 Omega;(ii) 0.3125 Omega`

Explanation

(i) `I=0.2 V^(5//2)implies0.44=0.2V^(5//2)`
`11/5=V^(5//2)implies121/25=V^(5)`
`V=1.37` volt
`I=0.2 V^(5//2)`
`I=(6-V)/R`
`0.44=(6-1.37)/Rimplies R=10.12 Omega`
(ii) `2I^(2)R=VIimplies2IR=V....(1)`
`V=(6-IR)....(2)`
From `(1)` and `(2)` `V=4` volt
`I=0.2 V^(5//2)=0.2 (4)^(5//2)=6.4`
`IR=2 implies6.4R=6`
`R=1/3.2=0.3125 Omega`