1.

Two identical 750-turn coils A and B lie in parallel plane. A current changing at the rate of 1500 A/s in A induces an emf of 11.25 V in B. (a)Calculate the mutual inductance of the arrangement.(b)If the inductance of each coil is 15mH, calculate the flux produced in coil A per ampere and, the percentage of this flux which links the turns of B.

Answer»



em = MdI1/dt

M = em/dI1/dt = 11.25/1500 = 7.5 * 10^-3 H = 7.5mH

Now,  L1 = (N1Φ1)/I1

So,  Φ1/I1 = L1/N1 = 15*10^-3/ 750 = 2* 10^-5Wb/A

Now, k= M/√L1L2 = (7.5 * 10^3)/√L2 = (7.5*10^3)/15*10^3

  = 0.5 Φ

= 50%

( since,  L1=L2=L)









 



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