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Two coils, A and B, have self inductances of 120μH and 300 μH respectively. A current of 1 A through coil A produces flux linkages of 100μWb turns in coil B. Calculate (i) the mutual inductance between the coils (ii) the coupling coefficient and (iii) the average e.m.f. induced in coil B if a current of 1 A in coil A is reversed at a uniform rate in 0.1 sec. |
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Answer» (i) M = (flux-linkages of coil B)/(current in coil A) = (100 x 10-6)/1 = 100 μH (ii) M = k√(L1L2) ∴ k = M/√(L1L2) = (100 x 10-6)/√(120 x 10-6 x 300 x 10-6) = 0.527 (iii) e2 = M × di/dt = (100 × 10−6) × 2/0.1 = 0.002 V or 2 mV. |
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