Current Electricity

JEE Main level, single correct, medium

165 of 380

Question 165 of 380

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

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Answer: (A) 4

Let resistances be R1R_1 and R2R_2
So, S=R1+R2S=R_1+R_2;
P=R1R2R1+R2 S=nPR1+R2=nR1R2R1+R2(R1+R2)2=nR1R2\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 R1=R2R_1=R_2, so minimum value of n=4n=4.

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