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Inductors - Engineering

Q1:

Five inductors are connected in series. The lowest value is 8 μH. If the value of each inductor is twice that of the preceding one, and if the inductors are connected in order of ascending values, the total inductance is

A 8 H

B 32 μH

C 64 μH

D 248 μH

ANS:D - 248 μH

Total inductance = L1 + L2 + L3 + L4 + L5.

L1 = 8 L2 = 16 L3 = 32 L4 = 64 L5 = 128.

8 + 16 + 32 + 64 + 128 = 248.