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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»
M = em/dI1/dt = 11.25/1500 = 7.5 * 10^-3 H = 7.5mHNow, L1 = (N1Φ1)/I1So, Φ1/I1 = L1/N1 = 15*10^-3/ 750 = 2* 10^-5Wb/ANow, 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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