Two inductors each of inductance $L$ are joined in parallel. Their equivalent inductance is

0

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2 L

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0.5 L

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L

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Solution

The correct option is C $0.5 L$
Parallel combination of inductor:
$\frac{1}{L_{e q u}} = \frac{1}{L_{1}} + \frac{1}{L_{2}}$
= $\frac{1}{L} + \frac{1}{L}$
= $\frac{2}{L}$
$∴ L_{e q u} = \frac{L}{2}$

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