Electrical Engineering - Inductors - Discussion

Discussion :: Inductors - General Questions (Q.No.20)

20. 

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

Answer: Option D

Explanation:

No answer description available for this question.

Prasanthi Rao said: (Sep 17, 2011)  
The inductances values are 8,16,32,64,128

Therefore total inductance connected in series = 8+16+32+64+128 = 248

Nt- said: (Oct 25, 2014)  
Or you could realize that you have a geometric series with 5 terms and use the formula.

s = a* (1-r^n) / (1-r) where in this case a is 8 uH, n is 5, and R is 2.

D.Ramesh said: (Dec 22, 2016)  
Total inductance = L1 + L2 + L3 + L4 + L5.

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

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

Srikanth said: (Jun 15, 2017)  
Yes, correct @D.Ramesh.

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