Two Pure inductors, each of self-inductance L are connected in parallel but are well
separated from each other, then the total inductance is

L

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2 L

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\frac{L}{2}

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\frac{L}{4}

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Solution

The correct option is C $\frac{L}{2}$
Equivalent inductance= $L$ (say)

$\frac{1}{L} = \frac{1}{L_{1}} + \frac{1}{L_{2}}$

$⟹ L = \frac{L_{1} L_{2}}{L_{1} + L_{2}}$

Here, equivalent inductance= $\frac{L . L}{L + L} = \frac{L}{2}$

Hence, answer is option-(C).

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