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A flux of 0.5 mWb is produced by a coil of 900 turns wound on a ring with a current of 3 A in it. Calculate (i) the inductance of the coil (ii) the e.m.f. induced in the coil when a current of 5 A is switched off, assuming the current to fall to zero in 1 milli second and (iii) the mutual inductance between the coils, if a second coil of 600 turns is uniformly wound over the first coil |
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Answer» (i) Inductance of the first coil = NΦ/I = (900 x 0.5 x 10-3)/3 = 0.15H (ii) e.m.f. induced e1 = L(di/dt) = 0.15 x (5 - 0)/(1 x 10-3) = 750V (iii) M(N2Φ1/I1) = (600 x 0.5 x 10-3)/3 = 0.1H (ii) Second Method for M We will now deduce an expression for coefficient of mutual inductance in terms of the dimensions of the two coils. Flux in the first coil Φ1 = N2I1/(l/μ0μrA) Wb; Flux/ampere = Φ1/I1 = N1/(l/μ0μrA) Assuming that whole of this flux (it usually is some percentage of it) is linked with the other coil having N2 turns, the weber-turns in it due to the flux/ampere in the first coil is M = N2Φ1/I1 = N2N1/(l/μ0μrA) ∴ M = ((μ0μrAN1N2)/l)H Also M = N1N2/(l/μ0μrA) = N1N2/reluctance = (N1N2/S)H |
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