The equivalent inductance of two dissimilar inductors is $2.4 H$ , when connected in parallel and $10 H$ , when connected in series. The inductance of one of the inductors is -

2 H

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6 H

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5 H

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7 H

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Solution

The correct option is B $6 H$
Let the inductances of the inductors be $L_{1}$ and $L_{2}$ .

If connected in series,

$L_{e q} = L_{1} + L_{2}$

$10 = L_{1} + L_{2} . . . (i)$

When connected in parallel,

$L_{e q} = \frac{L_{1} L_{2}}{L_{1} + L_{2}}$

$\Rightarrow 2.4 = \frac{L_{1} L_{2}}{L_{1} + L_{2}} . . . (i i)$

From equations $(i)$ and $(i i)$ ,

$L_{1} L_{2} = 24 . . . (i i i)$

Also,

$L_{1} - L_{2} = \sqrt{(L_{1} + L_{2})^{2} - 4 L_{1} L_{2}}$

$\Rightarrow L_{1} - L_{2} = \sqrt{10^{2} - 4 \times 24} [From (i) and (i i i)]$

$\Rightarrow L_{1} - L_{2} = 2 . . . (i v)$

On solving equations $(i)$ and $(i v)$ , we get, $L_{1} = 6 H$ and $L_{2} = 4 H$

Hence, option $(B)$ is the correct answer.

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The equivalent inductance of two dissimilar inductors is 2.4 H, when connected in parallel and 10 H, when connected in series. The inductance of one of the inductors is -

The equivalent inductance of two dissimilar inductors is $2.4 H$ , when connected in parallel and $10 H$ , when connected in series. The inductance of one of the inductors is -