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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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$5 Ω, 10 Ω$

$10 Ω, 5 Ω$

$5 Ω, 5 Ω$

$10 Ω, 10 Ω$

answer is D.

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Detailed Solution

(a) Given that $I = 0.4 A$

$∴ 0.4 = 0.2 V_{1}^{1 / 2} \Rightarrow V_{1} = 4 V,$

$V_{1} + V_{2} = 6 \Rightarrow V_{2} = 6 - V_{1} = 2 V$

$∴ R = V_{2} / I = 2 / 0.2 = 5 Ω$

$(b) P_{rod} = 2 P_{R}$

$\Rightarrow V_{1} I = 2 V_{2} I \Rightarrow V_{1} = 2 V_{2} \dots (1)$

$V_{1} + V_{2} = 6 \Rightarrow 2 V_{2} + V_{2} = 6 \Rightarrow V_{2} = 2 V$ and hence $V_{1} = 4 V$

$I = 0.2 V_{1}^{1 / 2} (in the rod) = 0.2 (4)^{1 / 2} = 0.4$

$= R = V_{2} / I = 2 / 0.4 = 5 Ω$

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