The resistance of the series combination of two resistance is $S$. When they are joined in parallel the total resistance is $P$. If $S = nP$, then the minimum possible value of $n$ is
AIEEE 2004,JEE MAIN 2021, Diffcult
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(a) If two resistances are ${R_1}$ and ${R_2}$ then
$S = {R_1} + {R_2}$ and $P = \frac{{{R_1}{R_2}}}{{({R_1} + {R_2})}}$
From given condition $S = nP$ i.e. $({R_1} + {R_2}) = n\,\left( {\frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}} \right)$
$==>$ ${({R_1} + {R_2})^2} = n\,\,{R_1}{R_2}$ $==>$ ${({R_1} - {R_2})^2} + 4{R_1}{R_2} = n{R_1}{R_2}$
So $n = 4 + \frac{{{{({R_1} - {R_2})}^2}}}{{{R_1}{R_2}}}.$ Hence minimum value of $n$ is $4$.
$S = {R_1} + {R_2}$ and $P = \frac{{{R_1}{R_2}}}{{({R_1} + {R_2})}}$
From given condition $S = nP$ i.e. $({R_1} + {R_2}) = n\,\left( {\frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}} \right)$
$==>$ ${({R_1} + {R_2})^2} = n\,\,{R_1}{R_2}$ $==>$ ${({R_1} - {R_2})^2} + 4{R_1}{R_2} = n{R_1}{R_2}$
So $n = 4 + \frac{{{{({R_1} - {R_2})}^2}}}{{{R_1}{R_2}}}.$ Hence minimum value of $n$ is $4$.
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