Basic Electrical Machines (8hrs) · Basics of Electrical & Electronics Engineering

mutual inductance

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mutual inductance

1. What Mutual Inductance Is

Mutual inductance (MM, henry HH) = property of two coils where changing current in one induces emf in the other via common flux:

e2=−MdI1/dt,e1=−MdI2/dte_2 = -M dI_1/dt , e_1 = -M dI_2/dt

where

M=N2φ21/I1=N1φ12/I2M = N_2 φ_{21}/I_1 = N_1 φ_{12}/I_2

(φ21φ_{21} = flux through coil 2 due to I1I_1). SI HH same as LL.

2. Coefficient of Coupling (kk)

Not all flux links both coils; coupling coefficient

k=M/√(L1L2)k = M/√(L_1 L_2)

where 0≤k≤10≤k≤1 (k=1k=1 perfect coupling all flux common, k=0k=0 no coupling). For tightly coupled (iron core transformer) k≈0.95k≈0.95–11; for air-core loose kk small.

Energy:

W=1/2L1I12+1/2L2I22±MI1I2W =1/2 L_1 I_1² +1/2 L_2 I_2² ± M I_1 I_2

(sign depends on dot convention).

3. Factors

M=k√(L1L2)=μN1N2A/lM = k√(L_1 L_2) = μ N_1 N_2 A / l

for solenoid sharing same core (ideal k=1k=1). So M∝N1N2M∝N_1 N_2, ∝μ∝μ, ∝A/l∝A/l, depends on geometry and separation.

4. Dot Convention

Dots on coils indicate reference polarity: if both currents enter dotted terminals, fluxes aid (+M+M); if one enters dot and other leaves, subtract. Determines sign of induced emf.

5. Worked Examples

(a) Two coils L1=0.4H, L2=0.6H, k=0.8 ⇒ M=0.8√(0.24)=0.39H ✓

(b) Transformer primary N1=100, N2=1000, same core l,A, μ ⇒ M=k·μ·N1·N2·A/l ⇒ voltage ratio N2/N1.

(c) Need to find induced emf: M=0.1H, dI1/dt=100A/s ⇒ e2= -10V ✓

(d) Why k<1? Leakage flux not linking both.

6. Exam Points

  • Define M=N2φ21/I1M=N_2 φ_{21}/I_1, e2=−MdI1/dte_2=-M dI_1/dt, HH — 5 marks, very frequent (almost certain).
  • Give k=M/√(L1L2)k=M/√(L1 L2) and 0≤k≤10≤k≤1 — 3-5 marks, frequent.
  • Mention dot convention — 3 marks, frequent.
  • Formula M=μN1N2A/lM=μ N1 N2 A/l for solenoid — 3-5 marks, frequent.
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