Electrical Engineering - Inductors - Discussion

Discussion :: Inductors - General Questions (Q.No.4)


A 5 mH, a 4.3 mH, and a 0.6 mH inductor are connected in parallel. The total inductance is

[A]. 9.9 mH
[B]. greater than 5 mH
[C]. 9.9 mH or greater than 5 mH
[D]. less than 0.6 mH

Answer: Option D


No answer description available for this question.

Satish Kumar said: (Jul 23, 2011)  
if the inductances(resistances) are connected in parallel,the equivalent inductances(resistances) will be less than the smaller individual inductance(resistance) coz... of the formula (1/Leq)=(1/L1)+(1/L2)+......+(1/Ln)

Thirumal said: (Dec 12, 2012)  
Ans : D
If the inductor are connected in parallel connection , then the formula for Equivalent Inductance is same as that of the Resistance.

i.e. (1/Leq)= (1/L1)+(1/L2)+(1/L3)

So (1/Leq) = (1/5m)+(1/4.3m)+(1/0.6m)
(1/Leq) = 2099.2248

Leq = 1/2099.2248 = 0.47 mH
Which is less than 0.6 mH.

Shivakumar K S said: (Dec 2, 2014)  
In any parallel combination of resistance or inductance the equivalent resistance or inductance is less than the least value of individual one.

Hence 0.6 is the least value in the combination of inductance in parallel.

ANS : 0.6.

Soujanya said: (Mar 28, 2018)  
Thank you @Shivakumar K S.

D Shwetha said: (Sep 8, 2018)  

Your explanation is short & good. Thank you.

Waseem said: (Oct 12, 2018)  
How to solve this? Please tell me.

Haseeb said: (Nov 16, 2020)  
The Equation. L = L2L3 + L1L3 + L1L2/L1L2L3.

Post your comments here:

Name *:

Email   : (optional)

» Your comments will be displayed only after manual approval.