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Q.

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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a

Power dissipated in R1 will be greater than the power dissipated in R2 if R1>R2.

b

Power dissipated in R1will be less than the power dissipated in R2 if R1

c

Power dissipated in R1will be equal to that in R2 for any value of R1 and R2.

d

Power dissipated in the whole network will be minimum.

answer is D.

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Detailed Solution

P=i12 R1+I22 R2=i12 R1+(i+i1)2 R2For minimum P,dpdl1=0. ⇒2i1R1-2(i-i1)R2=0 ⇒i1R1-i2R2=0 ⇒i1R1=i2R2But by ohm's Law, i1R1=i2R2 is always true.So power consumed by the system is always minimum.
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In the current carrying network shown in the figure, the current i distributes itself between the branches in such a way that