JEE MAIN - Physics (2022 - 27th June Evening Shift - No. 25)
In the given circuit 'a' is an arbitrary constant. The value of m for which the equivalent circuit resistance is minimum, will be $$\sqrt {{x \over 2}} $$. The value of x is __________.
Answer
3
Explanation
$${R_{net}} = {{ma} \over 3} + {a \over {2m}}$$
$$ = a\left[ {{m \over 3} + {1 \over {2m}} - {2 \over {\sqrt 6 }} + {2 \over {\sqrt 6 }}} \right]$$
$$ = a\left[ {{{\left( {\sqrt {{m \over 3}} - {1 \over {\sqrt {2m} }}} \right)}^2} + \sqrt {{2 \over 3}} } \right]$$
This will be minimum when
$$\sqrt {{m \over 3}} = {1 \over {\sqrt {2m} }}$$
or $$m = \sqrt {{3 \over 2}} $$
so $$x = 3$$
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