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The current I through a rod of a certain metallic oxide is given by I = 0.2 V^(1//2) , where V is the potential difference cross it. The rod is connected in series with a resistance to a 6 volt battery of negligible internal resistance. What value should the series resistance have so that : A) the current in the circuit is 0.4 A B) the power dissipated in the rod is twice that dissipated in the resistance. |
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Answer» `10 OMEGA, 10 Omega` |
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