The equivalent inductance of two inductors is $2.4 H$ when connected in parallel and 10 H when connected in series. What is the value of inductances of the individual inductors?

8 H, 2 H

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6 H, 4 H

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5 H, 5 H

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7 H, 3 H

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Solution

The correct option is B $6 H, 4 H$
In series connection $L_{1} + L_{2} = 10 H$ ...(i)
and in parallel connection $\frac{L_{1} L_{2}}{(L_{1} + L_{2})} = 2.4 H$ ...(ii)
Substituting the value of $(L_{1} + L_{2})$ from (i) into (ii), we get
$L_{1} L_{2} = (2.4) (L_{1} + L_{2}) = 2.4 \times 10 = 24$
$(L_{1} - L_{2})^{2} = (L_{1} + L_{2})^{2} - 4 L_{1} L_{2}$
$L_{1} L_{2} = [(10)^{2} - 4 \times 24]^{1 / 2} = 2 H$
Solving (i) and (ii), we get
$L_{1} = 6 H, L_{2} = 4 H$

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