JEE Mains · Physics · STD 12 - 3. current electricity
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 ...........

- A \(3\)
- B \(4\)
- C \(5\)
- D \(6\)
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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