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Power Delivered and Heat Dissipated in a Circuit

The inductor ...

Question

The inductor arrangement shown in the figure, with $L_{1} = 50 m H$ , $L_{2} = 80 m H$ , $L_{3} = 20 m H$ and $L_{4} = 15 m H$ , is to be connected to varying current source. What is the equivalent inductance of the arrangement?

18 m H

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55 m H

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71 m H

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81 m H

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Solution

The correct option is D $81 m H$
First the indutors $L_{2}$ and $L_{3}$ are in parallel. So equivalent inductance for these is $L_{23} = \frac{1}{1 / L_{2} + 1 / L_{3}} = \frac{L_{2} L_{3}}{L_{2} + L_{3}}$
Now $L_{1}, L_{23}$ and $L_{4}$ are in series. So the equivalent inductance of the circuit is
$L_{e q} = L_{1} + L_{23} + L_{4} = 50 + \frac{80 \times 20}{80 + 20} + 15 = 50 + 16 + 15 = 81 m H$

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