Current Electricity

JEE Main level, numerical answer, medium

141 of 380

Question 141 of 380

The equivalent resistance of series combination of two resistors is s.s. When they are connected in parallel, the equivalent resistance is p.p. If s=np,s=np, then the minimum value for nn is ______.
(Round off to the Nearest Integer)

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Answer: 4

R1+R2=s…(1){R}_{1}+{R}_{2}=s\ldots \left(1\right)
R1R2R1+R2=p…(2)\dfrac{{R}_{1}{R}_{2}}{{R}_{1}+{R}_{2}}=p\ldots \left(2\right)
R1R2=sp{R}_{1}{R}_{2}=sp
R1R2=np2{R}_{1}{R}_{2}=n{p}^{2}
R1+R2=nR1R2(R1+R2){R}_{1}+{R}_{2}=\dfrac{n{R}_{1}{R}_{2}}{\left({R}_{1}+{R}_{2}\right)}
(R1+R2)2R1R2=n\dfrac{{\left({R}_{1}+{R}_{2}\right)}^{2}}{{R}_{1}{R}_{2}}=n
for minimum value of nn
R1=R2=R{R}_{1}={R}_{2}=R
∴n=(2R)2R2=4∴n=\dfrac{(2R{)}^{2}}{{R}^{2}}=4

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