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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. |
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Answer» `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=5Omega` |
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