The equivalent inductance of two inductor is $2.4 m H$ when connected in parallel and $10 m H$ when connected in series. The difference between two inductance is (neglecting mutual induction between coils)

3 m H

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2 m H

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4 m H

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16 m H

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Solution

The correct option is C $2 m H$
The equivalent inductance of two inductors $L_{1}, L_{2}$ in series combination is
$L = L_{1} + L_{2}$ $= 10 H$ (given)
The equivalent inductance of two inductor $L_{1}, L_{2}$ in parallel combination is
$\frac{1}{L} = \frac{1}{L_{1}} + \frac{1}{L_{2}}$

$\frac{1}{2.4 H} = \frac{1}{L_{1}} + \frac{1}{L_{2}}$

$\frac{1}{2.4 H} = \frac{L_{1} + L_{2}}{L_{1} . L_{2}}$

$L_{1} . L_{2} = 10 \times 2.4 = 24$
$L_{1} = \frac{24}{L_{2}} H$

Put the value of $L 1$ in series combination, we get the quadratic equation,
$L_{2}^{2} - 12 L_{2} + 24 = 0$

$L_{2} = 4 H, 6 H$

$L_{1} = \frac{24}{4} H, \frac{24}{6} H$

$L_{1} = 6 H, 4 H$
The difference between the two inductance is $2 H$
$L_{1} = 6 H$
$L_{2} = 4 H$
$L_{1} - L_{2} = 2 H$

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