In the current carrying network shown in the figure, the current i distributes itself between the branches in such a way that

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Power dissipated in $R_{1}$ will be greater than the power dissipated in $R_{2} if R_{1} > R_{2} .$

Power dissipated in $R_{1}$ will be less than the power dissipated in $R_{2} if R_{1} < R_{2} .$

Power dissipated in $R_{1}$ will be equal to that in $R_{2}$ for any value of $R_{1} and R_{2}$ .

Power dissipated in the whole network will be minimum.

answer is D.

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Detailed Solution

$P = i_{1}^{2} R_{1} + I_{2}^{2} R_{2} = i_{1}^{2} R_{1} + {(i + i_{1})}^{2} R_{2}$
For minimum $P, \frac{dp}{{dl}_{1}} = 0 . \Rightarrow 2 i_{1} R_{1} - 2 (i - i_{1}) R_{2} = 0 \Rightarrow i_{1} R_{1} - i_{2} R_{2} = 0 \Rightarrow i_{1} R_{1} = i_{2} R_{2}$
But by ohm's Law, $i_{1} R_{1} = i_{2} R_{2}$ is always true.
So power consumed by the system is always minimum.