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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2022
  • Hard
  • Numerical

In the given circuit ' \(a\) ' is an arbitrary constant. The value of \(m\) for which the equivalent circuit resistance is minimum, will be \(\sqrt{\frac{ x }{2}}\). The value of \(x\) is ...........

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Answer & Solution

Correct Answer

(A) \(3\)

Step-by-step Solution

Detailed explanation

\(R =\left(\frac{ ma }{3}\right)+\left(\frac{ a }{2 m }\right)\)

\(\frac{ dR }{ dm }=\frac{ a }{3}-\frac{ a }{2 m ^{2}}=0\)

\(\frac{ a }{3}=\frac{ a }{2 m ^{2}}\)

\(m ^{2}=\frac{3}{2}\)

\(m =\sqrt{\frac{3}{2}}\)

\(x =3\)

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