# Electronics - Inductors - Discussion

### Discussion :: Inductors - General Questions (Q.No.58)

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

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.