Q196 / 225 · JEE Main 2004 · Physics

The resistance of the series combination of two resistances is $S$. When they are joined in parallel through total resistance is $P$. If $S=n P$, then the minimum possible value of $n$ is

  • Correct answer
    A
    4
  • B
    3
  • C
    2
  • D
    1

Solution

Let resistances be $R_1$ and $R_2$ So, $S=R_1+R_2$; $$ \begin{aligned} & \mathrm{P}=\frac{\mathrm{R}_1 \mathrm{R}_2}{\mathrm{R}_1+\mathrm{R}_2} \\ & \mathrm{~S}=\mathrm{nP} \\ & \mathrm{R}_1+\mathrm{R}_2=\frac{\mathrm{nR}_1 \mathrm{R}_2}{\mathrm{R}_1+\mathrm{R}_2} \\ & \left(\mathrm{R}_1+\mathrm{R}_2\right)^2=\mathrm{nR}_1 \mathrm{R}_2 \end{aligned} $$ If $R_1=R_2$, so minimum value of $n=4$.

Hence, the correct answer is option: 'A'.

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