Electronics - Inductors - Discussion

58. 

What is the total inductance of a 5 H and a 100 mH coil connected in parallel?

[A]. 4.76 mH
[B]. 33.3 mH
[C]. 98.0 mH
[D]. 150.0 mH

Answer: Option C

Explanation:

No answer description available for this question.

Clara said: (Jan 3, 2012)  
Inductors in parallel ==> 1/(1/L1 + 1/L2)

Dan said: (Feb 10, 2015)  
I don't understand:

I have divided 1/L1(1/5000 = 0.0002).

1/L2 (1/0.1 = 10).

Add together = 10.0002.

How do you get 98.0mH from this?

A previous similar question on parallel inductors I this section gave its answer as being 1/L1+1/L2+1/L3 = 1/600+1/300+1/800 = 0.00625.

TOTAL INDUCTANCE = 1/0.00625 = 160.

Why is the same principle not working here?

Tayler said: (Apr 8, 2015)  
Its because you are using 5000 in place of just 5 H. And 100 mH is 0.0100 H.

Try 1/(5^-1+0.0100^-1).

= 99 mH.

I can't get 98 mH from this but hey its pretty darn close ;).

If total inductance were 0.5 H and 100 mH you would get 98 mH as an answer.

Nithya said: (Jul 17, 2015)  
5H = 5000 mH.

Refer inductors in parallel formula mentioned above.

Using that substitute:

1/((1/5000)+(1/100)) mH = 98mH.

Abhishek Singh said: (Oct 2, 2016)  
1/LT = 1/L1 + 1/L2
LT = (L1 * L2)/(L11 + L2)

LT = (5000 * 100)/(5000 + 100).
= 500000/5100.
= 98.03.

Abhishek Singh said: (Oct 2, 2016)  
LT = (L1* L2)/(L1 + L2).

L1 = 5H = 5000mH.
L2 = 100mH.

LT = (5000 * 100)/(5000 + 100).
LT = 98.03mH.
LT = 98Mh.

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