JEE PYQ: Current Electricity Question 6
Question 6 - 2021 (17 Mar 2021 Shift 1)
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 of $n$ is (Round off to the Nearest Integer)
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Answer: 4
Solution
$R_1 + R_2 = s$ and $\frac{R_1 R_2}{R_1 + R_2} = p$. So $R_1 R_2 = sp = np^2$. Then $\frac{(R_1 + R_2)^2}{R_1 R_2} = n$. For minimum $n$, $R_1 = R_2 = R$: $n = \frac{(2R)^2}{R^2} = 4$.
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