---
title: "Parallel coupled inductors equivalent"
subject: "Basic Electrical Engineering"
chapter: 4
key: "semester_1/basic_electrical_engineering:mutual-inductance-parallel-equivalent"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/mutual-inductance-parallel-equivalent
site: BCT Engineering
---

# Parallel coupled inductors equivalent

Basic Electrical Engineering · Chapter 4: Inductance & Capacitance in Electric Circuits

- Asked in: 2074 Asoj, 4 marks; 2069 Asar, 8 marks (asked 2 times)
- Video: [Derivation of equivalent inductance of two inductors connected in parallel • HERO OF THE DERIVATIONS](https://www.youtube.com/watch?v=6MDYVvAnvbM) — Hero of the derivations
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/mutual-inductance-parallel-equivalent
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-4

## Question

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?

## 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.

## Answer

#### 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.

```text
               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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In the aiding case, how are the coil equations written?
   A. v = (L1 + L2) di/dt
   B. v = L1 di1/dt + M di2/dt and v = L2 di2/dt + M di1/dt
   C. v = L1 di1/dt - M di2/dt
   D. v = M di/dt
   Answer: B. Each coil has the same voltage and includes a mutual term driven by the other coil's current. (Key point: Equations (aiding))

2. What is L_eq for aiding flux?
   A. (L1L2 - M^2)/(L1 + L2 - 2M)
   B. (L1L2 + M^2)/(L1 + L2 + 2M)
   C. L1L2/(L1 + L2 - 2M)
   D. (L1L2 - M^2)/(L1 + L2 + 2M)
   Answer: A. Adding di1/dt and di2/dt gives di/dt = v(L1 + L2 - 2M)/(L1L2 - M^2), so L_eq is the inverse factor. (Key point: Derived result)

3. How is the opposing result obtained from the aiding result?
   A. Set M = 0
   B. Replace L by -L
   C. Swap L1 and L2
   D. Replace M by -M, giving (L1L2 - M^2)/(L1 + L2 + 2M)
   Answer: D. Opposing flux reverses the sign of M. (Key point: Opposing case)

4. For L1 = 4 H, L2 = 6 H, M = 3 H, what are aiding and opposing values?
   A. 3.75 H and 0.9375 H
   B. 2.4 H and 3.75 H
   C. 15 H and 3.75 H
   D. 0.9375 H and 3.75 H
   Answer: A. 15/4 = 3.75 H aiding and 15/16 = 0.9375 H opposing. (Key point: Example)

## Related in Basic Electrical Engineering

- [Equivalent inductance of two coupled coils in parallel](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Acoupled-coils-4h-6h-parallel-3h) — Basic Electrical Engineering, chapter 4
- [Inductance, series mutual flux aiding](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Ainductance-series-aiding) — Basic Electrical Engineering, chapter 4
- [Inductance and inductors in parallel](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Ainductors-in-parallel-derivation) — Basic Electrical Engineering, chapter 4, asked 2×
- [Inductance from flux linkages](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Ainductance-flux-linkages-derivation) — Basic Electrical Engineering, chapter 4

## Related in other subjects

- [CFGs for L1, L2; L1∩L2 not CF](https://www.bctengineer.com/question.md?key=semester_3%2Ftheory_of_computation%3Acfg-l1-l2-pumping-proof) — Theory of Computation, chapter 3
- [Equivalent Y-parameter of parallel two-ports](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aequivalent-y-parameter-parallel-two-ports) — Electrical Circuit Theory, chapter 7, asked 3×
- [Equivalent ABCD of lines in parallel](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aparallel-lines-abcd-derivation) — Power System Analysis I, chapter 5
- [Equivalent T from Y parameters of series two-ports](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aequivalent-t-parameter-two-y-networks) — Electrical Circuit Theory, chapter 7