Discussion :: Inductors  General Questions (Q.No.10)
10. 
What is the total inductance in the given circuit? 

Answer: Option A Explanation: No answer description available for this question.

Gjhoj said: (Nov 9, 2011)  
1/600+1/300+1/800 .006+.003+.008 
Tejas Savani said: (Apr 6, 2012)  
Total= 1/[(1/L1) + (1/L2) + (1/L3)] = 1/0.00625 = 160 mH 
Diwakar said: (Apr 23, 2012)  
1/Ltotal =1/L1+1/L2+1/L3 =1/600+1/300+1/800 =15/2400 L =2400/15 =160 
Jayashri said: (Feb 5, 2014)  
How comes that answer, please tel me I have doubt why are you taken 15 number in calculation. 
Jeya said: (Feb 20, 2014)  
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. 
Anjali said: (Apr 24, 2016)  
Why are you taking 15 number for calculation? 
Shaheer said: (Aug 25, 2016)  
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 
Sanjay said: (Dec 2, 2016)  
Without mutual induction magnitude. How this could be possible? 
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