--- title: "Inductance and inductors in parallel" subject: "Basic Electrical Engineering" chapter: 4 key: "semester_1/basic_electrical_engineering:inductors-in-parallel-derivation" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/inductors-in-parallel-derivation site: BCT Engineering --- # Inductance and inductors in parallel Basic Electrical Engineering · Chapter 4: Inductance & Capacitance in Electric Circuits - Asked in: 2069 Chaitra, 4 marks; 2067 Asoj, 8 marks (asked 2 times) - Video: [Inductors in Parallel, Equivalent Inductance](https://www.youtube.com/watch?v=qeQjMhAaLYA) — eeKnowHow - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/inductors-in-parallel-derivation - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-4 ## Question Define inductance and derive relation for connection of inductors connected in parallel connection. ## Short answer Inductance is the property of a coil opposing current change, defined as flux linkage per ampere (L = NΦ/I). In parallel without mutual coupling, voltage is same across all, so current change rates sum; therefore 1/L_eq = 1/L1 + 1/L2 + 1/L3. ## Answer #### Inductance **Inductance (L)** is the property of a coil by which it opposes any change of current through it. It is the flux linkage per ampere: L = NΦ/I (henry) From Faraday's law, the induced EMF is e = −L di/dt, so L is the EMF induced per unit rate of change of current. 1 henry produces 1 volt when the current changes at 1 A/s. #### Inductors in parallel (no mutual coupling) ```text i1 → +---(L1)---+ i → | i2 → | o---------+---(L2)---+---------o | i3 → | +---(L3)---+ ``` All inductors have the same voltage *v* across them, and the supply current divides: i = i₁ + i₂ + i₃. For each inductor, v = L di/dt, so di₁/dt = v/L₁, di₂/dt = v/L₂, di₃/dt = v/L₃ Differentiate the current equation: di/dt = di₁/dt + di₂/dt + di₃/dt = v (1/L₁ + 1/L₂ + 1/L₃) For the equivalent inductor, di/dt = v/L_eq, so **1/L_eq = 1/L₁ + 1/L₂ + 1/L₃** For two inductors: L_eq = L₁L₂/(L₁ + L₂). #### Notes - L_eq is smaller than the smallest inductance. - The current divides inversely to the inductances: i₁ = i × L₂/(L₁ + L₂). - Example: 6 H and 3 H in parallel give 2 H. - If the coils are magnetically coupled with mutual inductance M, L_eq = (L₁L₂ − M²)/(L₁ + L₂ ∓ 2M) (minus for aiding flux, plus for opposing). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is inductance? A. The opposition to steady current B. The flux per unit voltage C. The ability to store charge D. The property opposing change of current, L = N phi/I Answer: D. e = -L di/dt; 1 H gives 1 V at 1 A/s. (Key point: Inductance) 2. What is common to inductors in parallel? A. The inductance B. The current through each C. The flux D. The voltage across them Answer: D. All share the same voltage; the supply current divides. (Key point: Parallel basis) 3. What is derived? A. 1/L_eq = L1 + L2 + L3 B. 1/L_eq = 1/L1 + 1/L2 + 1/L3 C. L_eq = L1L2L3 D. L_eq = L1 + L2 + L3 Answer: B. Differentiating i = i1 + i2 + i3 with di/dt = v/L gives the reciprocal sum. (Key point: Result) 4. With coupling M and aiding flux, how does the formula change? A. L_eq = (L1L2 - M^2)/(L1 + L2 - 2M) B. L_eq unchanged C. L_eq = L1L2/(L1 + L2 + 2M) D. L_eq = L1 + L2 - 2M Answer: A. The uncoupled parallel formula is modified by M; aiding uses minus 2M in the denominator. (Key point: Current division and coupling) ## Related in Basic Electrical Engineering - [Parallel coupled inductors equivalent](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Amutual-inductance-parallel-equivalent) — Basic Electrical Engineering, chapter 4, asked 2× - [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 - [Delta connected coils line current and power](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Adelta-coils-7ohm-0-03h-400v) — Basic Electrical Engineering, chapter 8 - [Parallel RL and RC branches](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aparallel-rl-rc-240v-30ohm-75mh) — Basic Electrical Engineering, chapter 6, asked 2× - [Parallel RL and RC branches, total current](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aparallel-rl-rc-total-current-pf-power) — Basic Electrical Engineering, chapter 6, asked 2× - [Coil on 240 V, power, current, inductance](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Acoil-1-3kva-1-2kvar-240v) — Basic Electrical Engineering, chapter 6 ## Related in other subjects - [Parallel connection of two-port networks](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atwo-port-parallel-connection) — Filter Design, chapter 4 - [Derive Mohr-Coulomb principal stress relation](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Amohr-coulomb-derive-principal-stresses) — Soil Mechanics, chapter 10, asked 4× - [Derive parallel projection matrix](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Aparallel-projection-matrix-derive) — Computer Graphics, chapter 4, asked 2× - [DLG fault sequence network connection](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Adlg-sequence-network-connection) — Power System Analysis II, chapter 5, asked 3×