Electronics - Inductors - Discussion
Discussion Forum : Inductors - General Questions (Q.No. 10)
10.
What is the total inductance in the given circuit?
Discussion:
9 comments Page 1 of 1.
Gjhoj said:
1 decade ago
1/600+1/300+1/800
.006+.003+.008
.006+.003+.008
Tejas Savani said:
1 decade ago
Total= 1/[(1/L1) + (1/L2) + (1/L3)]
= 1/0.00625
= 160 mH
= 1/0.00625
= 160 mH
Diwakar said:
1 decade ago
1/Ltotal =1/L1+1/L2+1/L3
=1/600+1/300+1/800
=15/2400
L =2400/15
=160
=1/600+1/300+1/800
=15/2400
L =2400/15
=160
Jayashri said:
1 decade ago
How comes that answer, please tel me I have doubt why are you taken 15 number in calculation.
Jeya said:
1 decade ago
Hi @Jayashri 15 is nothing but just he taken LCM.
1/Ltotal = 1/L1+1/L2+1/L3.
= 1/300+1/600+1/800.
LCM of denominator is 2400.
Above equation becomes,
1/Ltotal = (8+4+3)/2400 = 15/2400.
1/Ltotal = 15/2400.
Ltotal = 2400/15 = 160.
1/Ltotal = 1/L1+1/L2+1/L3.
= 1/300+1/600+1/800.
LCM of denominator is 2400.
Above equation becomes,
1/Ltotal = (8+4+3)/2400 = 15/2400.
1/Ltotal = 15/2400.
Ltotal = 2400/15 = 160.
Anjali said:
9 years ago
Why are you taking 15 number for calculation?
Shaheer said:
9 years ago
Hi @Anjali.
Formulae for inductance L.
1/L = 1/L1 + 1/L2 + 1/L3.
Therefore substituting the given values in formulae,
1/L = 1/600 + 1/300 + 1/800.
L.C.M of 600, 300, 800 is 2400.
SO
1/L = 4/2400 + 8/2400 + 3/2400.
1/L = 15/2400.
THEREFORE L = 2400/15.
= 160 -> Answer
Formulae for inductance L.
1/L = 1/L1 + 1/L2 + 1/L3.
Therefore substituting the given values in formulae,
1/L = 1/600 + 1/300 + 1/800.
L.C.M of 600, 300, 800 is 2400.
SO
1/L = 4/2400 + 8/2400 + 3/2400.
1/L = 15/2400.
THEREFORE L = 2400/15.
= 160 -> Answer
(3)
Sanjay said:
9 years ago
Without mutual induction magnitude. How this could be possible?
Rupali said:
2 years ago
How comes that answer, please explain me.
@Shaheer.
Here why are you taking 15 in the calculation?
@Shaheer.
Here why are you taking 15 in the calculation?
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