If the ratio of the equivalent resistance of | vedclass

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(D) The equivalent resistance in series is $R_S = R_1 + R_2$.The equivalent resistance in parallel is $R_P = \frac{R_1 R_2}{R_1 + R_2}$.Given the rati

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${\left( {\frac{{{R_1}}}{{{R_2}}}} \right)^{1/2}} + {\left( {\frac{{{R_2}}}{{{R_1}}}} \right)^{1/2}} = {n^{1/2}}$

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(D) The equivalent resistance in series is $R_S = R_1 + R_2$.The equivalent resistance in parallel is $R_P = \frac{R_1 R_2}{R_1 + R_2}$.Given the ratio $\frac{R_S}{R_P} = n$,we have:$\frac{R_1 + R_2}{\frac{R_1 R_2}{R_1 + R_2}} = n$This simplifies to:$\frac{(R_1 + R_2)^2}{R_1 R_2} = n$Taking the square root on both sides:$\frac{R_1 + R_2}{\sqrt{R_1 R_2}} = \sqrt{n}$Dividing each term in the numerator by the denominator:$\frac{R_1}{\sqrt{R_1 R_2}} + \frac{R_2}{\sqrt{R_1 R_2}} = \sqrt{n}$This results in:$\sqrt{\frac{R_1}{R_2}} + \sqrt{\frac{R_2}{R_1}} = \sqrt{n}$Which can be written as:${\left( {\frac{{{R_1}}}{{{R_2}}}} \right)^{1/2}} + {\left( {\frac{{{R_2}}}{{{R_1}}}} \right)^{1/2}} = {n^{1/2}}$.

If the ratio of the equivalent resistance of two resistors $R_1$ and $R_2$ in series to that in parallel is $n$,then:

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