Electrical Engineering - Inductors - Discussion
Discussion Forum : Inductors - General Questions (Q.No. 22)
22.
The following coils are in parallel: 75 
H, 40 H, 25 H, and 15 H. The total inductance is approximately
H, 40 H, 25 H, and 15 H. The total inductance is approximately6.9 H
H14 H
H2.2 H
H155 H
HDiscussion:
7 comments Page 1 of 1.
Hemasundar said:
4 years ago
Thanks for the explanation.
(1)
Arjun said:
9 years ago
75||40 = 26.08.
25||15 = 9.375.
26.08||9.375 = 6.9.
25||15 = 9.375.
26.08||9.375 = 6.9.
(1)
Mukhtar said:
10 years ago
Same formula as that of the resistors when connected in parallel.
ARUN KUMAR M said:
1 decade ago
1/L = 1/L1+1/L2+1/L3+1/L4.
1/L = 1/75+1/40+1/25+1/15.
1/L = 0.013+0.025+0.04+0.066.
1/L = 0.144.
L = 1/0.144.
L = 6.9.
1/L = 1/75+1/40+1/25+1/15.
1/L = 0.013+0.025+0.04+0.066.
1/L = 0.144.
L = 1/0.144.
L = 6.9.
(2)
Parneet kaur said:
1 decade ago
Can anybody solve this sum in detail?
Biresh said:
1 decade ago
1/75+1/40+1/25+1/15
Vandan said:
1 decade ago
Total inductance = Lp
and
1/Lp=1/L1 + 1/L2 + 1/L3 + 1/L4
Here L1, L2, L3, L4 are 75, 40, 25, 15.
and
1/Lp=1/L1 + 1/L2 + 1/L3 + 1/L4
Here L1, L2, L3, L4 are 75, 40, 25, 15.
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