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The total inductance of two coils, A and B, when connected in series, is 0.5 H or 0.2 H, depending on the relative directions of the current in the coils. Coil A, when isolated from coil B, has a self-inductance of 0.2 H. Calculate (a) the mutual inductance between the two coils (b) the self-inductance of coil B (c) the coupling factor between the coils. (d) the two possible values of the induced e.m.f. in coil A when the current is decreasing at 1000 A per second in the series circuit. |
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Answer» (a) Combined inductance is given by L = L1 + L2 ± 2M ∴ 0.5 = L1 + L2 + 2M ...(i), 0.2 = L1 + L2 − 2M ...(ii) Subtracting (ii) from (i), we have 4M = 0.3 or M = 0.075 H (b) Adding (i) and (ii) we have 0.7 = 2 × 0.2 + 2L2 = 0.15 H (c) Coupling factor or coefficient is k = M/√(L1L2) = 0.075/√(0.2 + 0.15) = 0.433 or 43.4% (d) e1 = L1(di/dt) ± M(di/dt) ∴ e1 = (0.2 + 0.075) × 1000 = 275 V = (0.2 − 0.075) × 1000 = 125 V |
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