1.two coupled coils with L1=L2=0.6 H have a coupling
coefficient of k=0.8 the turns ratio N1/N2 is......
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Answer / ratnakar
as per lenz law
E1=-L1d(i)/dt
E2=-L2d(i)/dt
E1/E2=L1/L2=N1/N1=0.6/0.6=1
this question is asked in IES 2013 hope my answer is clear
Is This Answer Correct ? | 75 Yes | 8 No |
Answer / achintya kumar shit
We know that, Inductance L=(N*phai)/l
So, L1/L2=N1/N2
or, 0.6/0.6=N1/N2
or, N1/N2=1
Is This Answer Correct ? | 32 Yes | 4 No |
Answer / shekhar
L proportional to N^2
Since L1=L2=0.6
We can assume, N1^2=N2^2=(0.6)^2
Hence, N1/N2=1.
Is This Answer Correct ? | 3 Yes | 1 No |
Answer / vinova
well frnd it cant be calculated as you can calculate mutual
inductance M = K*underoot(L1*L2)
Further M=(N1*N2)/S (s is reluctance)so the data in the
above question is insufficent to calculate N1 and N2. if it
is asssumed that both have same length l and both carry
same current and both have equal magnatic field intensity H
then it can be said that N1 and N2 are equal as H*l = N*I
here l s the length of conductor
Is This Answer Correct ? | 2 Yes | 4 No |
coupling coefficient k=n2/n1
so, n2/n1 =1/k=1/0.8 = 1.25
Is This Answer Correct ? | 3 Yes | 23 No |
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