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Parallel coupled inductors equivalent

Short answer

For aiding mutual flux, equivalent inductance Leq = (L1L2 – M²)/(L1 + L2 – 2M). For opposing flux, Leq = (L1L2 – M²)/(L1 + L2 + 2M). Derived from v = L1 di1/dt ± M di2/dt and v = L2 di2/dt ± M di1/dt, solving for total di/dt.

Two inductances L1 and L2 are connected in parallel. Derive the relation showing the equivalent inductance of the combination when mutual flux helps the individual flux. What will be the equivalent inductance of the combination when mutual flux opposes the individual flux?

Derivation (mutual flux aiding)

Two coils L₁ and L₂ with mutual inductance M are connected in parallel, with currents i₁ and i₂, so the supply current is i = i₁ + i₂. The same voltage v appears across both coils.

               i1 →
           +---(L1)---+
   i →     |    ↕ M   |
 o---------+          +---------o
           |   i2 →   |
           +---(L2)---+

When the mutual flux helps the self flux (aiding):

v = L₁ di₁/dt + M di₂/dt ... (1) v = L₂ di₂/dt + M di₁/dt ... (2)

Solving (1) and (2) for the two rates of change:

di₁/dt = v (L₂ − M)/(L₁L₂ − M²) di₂/dt = v (L₁ − M)/(L₁L₂ − M²)

Adding:

di/dt = di₁/dt + di₂/dt = v (L₁ + L₂ − 2M)/(L₁L₂ − M²)

For an equivalent inductance, v = L_eq di/dt, so

L_eq (aiding) = (L₁L₂ − M²)/(L₁ + L₂ − 2M)

Mutual flux opposing

When the mutual flux opposes the self flux, M changes sign (−M):

L_eq (opposing) = (L₁L₂ − M²)/(L₁ + L₂ + 2M)

Special case and example

  • If M = 0, L_eq = L₁L₂/(L₁ + L₂), the usual parallel formula.
  • Example: L₁ = 4 H, L₂ = 6 H, M = 3 H: aiding: (24 − 9)/(10 − 6) = 3.75 H opposing: (24 − 9)/(10 + 6) = 0.9375 H

✓ Answer checked by Claude Opus 5.5 (high effort)

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