Four resistors, 100 Omega, 200 Omega, 300 Ome | vedclass

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(D) To maximize the equivalent resistance across a diagonal, we must group the resistors such that the two parallel branches have the largest possible

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$250$

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(D) To maximize the equivalent resistance across a diagonal, we must group the resistors such that the two parallel branches have the largest possible sum of resistances. Let the four resistors be $R_1, R_2, R_3, R_4$. When connected across a diagonal, they form two parallel branches, each containing two resistors in series. Let the branches be $(R_a + R_b)$ and $(R_c + R_d)$. The equivalent resistance is $R_{eq} = \frac{(R_a + R_b)(R_c + R_d)}{(R_a + R_b) + (R_c + R_d)}$. To maximize this, we should make the sums of the two branches as close to each other as possible. The total sum is $100 + 200 + 300 + 400 = 1000 \Omega$. Thus, we aim for each branch to be $500 \Omega$. We can pair them as $(400 + 100) = 500 \Omega$ and $(300 + 200) = 500 \Omega$. The equivalent resistance is then $\frac{500 	imes 500}{500 + 500} = \frac{250000}{1000} = 250 \Omega$.

Four resistors, $100 \Omega, 200 \Omega, 300 \Omega$ and $400 \Omega$ are connected to form four sides of a square. The resistors can be connected in any order. What is the maximum possible equivalent resistance across the diagonal of the square (in $\Omega$)?

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