Two resistances R1 and R2 provide series to parallel equivalents as n-1- Then the correct relationship is

Series resistance R_S = R_1 + R_2 and parallel resistance R_P = R_1 R_2/R_1 + R_2⇒R_S/R_P = (R_1+ R_2)^2/R_1 R_2 = n ⇒R_1 + R_2/√(R_1 R_2)=√(n)⇒√(R_1^2)/√(R_ ...

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Byju's Answer

Standard XII

Physics

Series and Parallel Combination of Resistors

Two resistanc...

Question

Two resistances $R_{1}$ and $R_{2}$ provide series to parallel equivalents as $\frac{n}{1}$ . Then the correct relationship is

{(\frac{R_{1}}{R_{2}})}^{2} + {(\frac{R_{2}}{R_{1}})}^{2} = n^{2}

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{(\frac{R_{1}}{R_{2}})}^{3 / 2} + {(\frac{R_{2}}{R_{1}})}^{3 / 2} = n^{3 / 2}

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(\frac{R_{1}}{R_{2}}) + (\frac{R_{2}}{R_{1}}) = n

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{(\frac{R_{1}}{R_{2}})}^{1 / 2} + {(\frac{R_{2}}{R_{1}})}^{1 / 2} = n^{1 / 2}

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Solution

The correct option is D ${(\frac{R_{1}}{R_{2}})}^{1 / 2} + {(\frac{R_{2}}{R_{1}})}^{1 / 2} = n^{1 / 2}$
Series resistance $R_{S} = R_{1} + R_{2}$ and parallel resistance $R_{P} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} \Rightarrow \frac{R_{S}}{R_{P}} = \frac{(R_{1} + R_{2})^{2}}{R_{1} R_{2}} = n$
$\Rightarrow \frac{R_{1} + R_{2}}{\sqrt{R_{1} R_{2}}} = \sqrt{n} \Rightarrow \frac{\sqrt{R_{1}^{2}}}{\sqrt{R_{1} R_{2}}} + \frac{\sqrt{R_{2}^{2}}}{\sqrt{R_{1} R_{2}}} = \sqrt{n} \Rightarrow \sqrt{\frac{R_{1}}{R_{2}}} + \sqrt{\frac{R_{2}}{R_{1}}} = \sqrt{n}$

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Two resistances R1 and R2 provide series to parallel equivalents as n1. Then the correct relationship is

The correct option is D (R1R2)1/2+(R2R1)1/2=n1/2Series resistance RS=R1+R2 and parallel resistance RP=R1R2R1+R2⇒RSRP=(R1+R2)2R1R2=n ⇒R1+R2√R1R2=√n⇒√R21√R1R2+√R22√R1R2=√n⇒√R1R2+√R2R1=√n

Two resistances $R_{1}$ and $R_{2}$ provide series to parallel equivalents as $\frac{n}{1}$ . Then the correct relationship is