JEE MAIN - Physics (2021 - 17th March Morning Shift - No. 26)
The equivalent resistance of series combination of two resistors is 's'. When they are connected in parallel, the equivalent resistance is 'p'. If s = np, then the minimum value for n is ____________. (Round off to the Nearest Integer)
Answer
4
Explanation
$$s = np$$
$${R_1} + {R_2} = n\left[ {{{{R_1}{R_2}} \over {{R_1} + {R_2}}}} \right]$$
$$ \Rightarrow $$ $$R_1^2 + R_2^2 + 2{R_1}{R_2} = n{R_1}{R_2}$$
$$ \Rightarrow $$ $$R_1^2 + (2 - n){R_1}{R_2} + R_2^2 = 0$$
For real roots, b2 - 4ac $$ \ge $$ 0
$${[(2 - n){R_2}]^2} - 4 \times 1 \times R_2^2$$ $$ \ge $$ 0
$$ \Rightarrow $$ $${(2 - 4)^2}R_2^2 \ge 4R_2^2$$
$$ \Rightarrow $$ 2 $$-$$ n $$\ge $$$$\pm$$2
$$ \Rightarrow $$ 2 $$-$$ n $$\ge$$ $$-$$2
$$ \Rightarrow $$ n $$\ge$$ 4
So, minimum value for n = 4
$${R_1} + {R_2} = n\left[ {{{{R_1}{R_2}} \over {{R_1} + {R_2}}}} \right]$$
$$ \Rightarrow $$ $$R_1^2 + R_2^2 + 2{R_1}{R_2} = n{R_1}{R_2}$$
$$ \Rightarrow $$ $$R_1^2 + (2 - n){R_1}{R_2} + R_2^2 = 0$$
For real roots, b2 - 4ac $$ \ge $$ 0
$${[(2 - n){R_2}]^2} - 4 \times 1 \times R_2^2$$ $$ \ge $$ 0
$$ \Rightarrow $$ $${(2 - 4)^2}R_2^2 \ge 4R_2^2$$
$$ \Rightarrow $$ 2 $$-$$ n $$\ge $$$$\pm$$2
$$ \Rightarrow $$ 2 $$-$$ n $$\ge$$ $$-$$2
$$ \Rightarrow $$ n $$\ge$$ 4
So, minimum value for n = 4
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