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Question

Two inductors, each of inductance L are connected in parallel. One more inductor of inductance 5 mH is connected in series with this combination. Now, the effective inductance is 15 mH. The value of L is

A
10 mH
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B
50 mH
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C
15 mH
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D
20 mH
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Solution

The correct option is D 20 mH
The given configuration is shown in the figure.

For inductors between points A and B, net inductance is,

Leq=L1L2L1+L2=L×LL+L=L2

The configuration reduced into the below shown circuit part.


For inductors between points A and C, net inductance,

Leq=L1+L2

15=L2+5

L=20 mH

Hence, option (D) is the correct answer.

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