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 __________.

JEE Main 2022 (Online) 27th June Evening Shift Physics - Current Electricity Question 149 English

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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