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Two inductors A and B when connected in parallel are equivalent to a single inductor of inductance 1.5 H, and when connected in series are equivalent to a single inductor of inductance 8 H. Find the difference in the inductances of A and B.

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3 H

7.5 H

2 H

4 H

answer is D.

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Detailed Solution

$\frac{L_{1} L_{2}}{L_{1} + L_{2}} = 1.5 L_{1} + L_{2} = 8 {(L_{1} - L_{2})}^{2} = {(L_{1} + L_{2})}^{2} - 4 L_{1} L_{2} ∴ L_{1} - L_{2} = 4 H$

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