Electrical Engineering - Transformers - Discussion
Discussion Forum : Transformers - General Questions (Q.No. 8)
8.
The mutual inductance when k = 0.65, L1 = 2 
H, and L2 = 5 H is
H, and L2 = 5 H is2 mH
2 H
H4 H
H8 H
HDiscussion:
39 comments Page 3 of 4.
Pramod rps said:
1 decade ago
Its micro obviously.
Priyanka said:
1 decade ago
L2=5, not 10 hows write 10.
P venkateswara reddy said:
10 years ago
As we know that M = K (L1*L2)^(1/2).
M = 0.65*(2*5*10^-6*10^-6)^(1/2) = 2 uH.
M = 0.65*(2*5*10^-6*10^-6)^(1/2) = 2 uH.
NAKUL said:
10 years ago
I can't understand this example, please explain.
Sravanthi said:
9 years ago
The correct formulae are k=M/(L1*L2)^1/2.
Shekar said:
9 years ago
M = k * (L1 L2)^1/2 .
Jittu said:
9 years ago
Good answer. Thanks you all.
Manjunath Ukkali said:
9 years ago
Not getting it. Please help me.
(1)
P. RAJESH said:
8 years ago
How to apply 5*10 in sum?
(2)
Shenbalaxmi said:
8 years ago
K-Coupling factor
L1,L2-Self inductance
M-Mutual inductance
M = K * sqrt(L1 * L2),
M = 0.65 * (2x10^-6x5x10^-6).
M = 2microHenry.
L1,L2-Self inductance
M-Mutual inductance
M = K * sqrt(L1 * L2),
M = 0.65 * (2x10^-6x5x10^-6).
M = 2microHenry.
(6)
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