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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Current Electricity
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The resistance of the series combination of two resistances is $S$. When they are joined in parallel through total resistance is $P$. If $S = nP$, then the minimum possible value of $n$ is

$\begin {array} {1 1} (A)\;1 & \quad (B)\;2 \\ (C)\;3 & \quad (D)\;4 \end {array}$


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Let the resistances be $ R_1\: and\: R_2$
So $ S = R_1 + R_2$
$ P = \large\frac{R_1R_2}{R_1+R_2}$
$ S = nP$
$ R_1+R_2 = \large\frac{nR_1R_2}{R_1+R_2}$
$( R_1+R_2)^2=nR_1R_2$
Minimum occurs for $R_1 = R_2 $ and the minimum value is $4$
Ans : (D)


answered Feb 16, 2014 by thanvigandhi_1
edited Mar 14, 2014 by thanvigandhi_1

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