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Inductors

Q1: The total inductance in the given circuit is 0.6 mH.

A True

B False

ANS:A - True

LT = L1 + L2 + L3 + (For Inductors connected in SERIES).
1/( LT ) = 1/( L1 ) + 1/( L2 ) + 1/( L3 ) +.. (For Inductors connected in PARALLEL).

Here: L2 and L5 are in SERIES.
LT1 = L2 + L5.
LT1 = (300 * 10^ (-6)) + (500 * 10^ (-6)).
LT1 = (800 * 10^ (-6)) = 800m H.

Now the resultant Inductance LT1 = 800mH is in PARALLEL with SERIES combination of (L3 and L4).
LT2 = L3 + L4.
LT2 = (100 * 10^ (-6)) + (700 * 10^ (-6)).
LT2 = (800 * 10^ (-6)) = 800 mH.

1/ (LT3) = 1/ (LT1) + 1/ (LT2).
1/ (LT3) = [1 / (800 * 10^ (-6))] + [1 / (800 * 10^ (-6))].
1/ (LT3) = (0.0025 * 10^ (6)).
LT3 = [1 / (0.0025 * 10^ (6))].
LT3 = (400 * 10^ (-6)) = 400 mH.

The resultant Inductance LT3= 400mH is in SERIES with L1.
LT4 = LT3 + L1.
LT4 = (400 * 10^ (-6)) + (200 * 10^ (-6)).
LT4 = (6 00 * 10^ (-6)) = 600mH.
LT4 = 600mH = 0.60 mH.



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