---
title: "Coil current for 2 mWb in central limb of core"
subject: "Electrical Machines I"
chapter: 1
key: "semester_4/electrical_machines_i:three-limb-two-coils-central-core"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/three-limb-two-coils-central-core
site: BCT Engineering
---

# Coil current for 2 mWb in central limb of core

Electrical Machines I · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2068 Bhadra, 8 marks
- Video: [ANALYSIS OF PARALLEL MAGNETIC CIRCUITS   1](https://www.youtube.com/watch?v=lCyiphcg-20) — Archana Manuel
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/three-limb-two-coils-central-core
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-1

## Question

For magnetic circuit shown in figure below, calculate the current to be passed through coil A so that magnetic flux in the central core is 2 mWb. Given that the relative permeability of core = 1000. Given that: I2 = 5 Amp, A1 = 4 cm², A2 = 2 cm², AB = CD = EF = 15 cm, BC = AD = BE = AF = 15 cm. [Figure: three-limb core with top yoke E–B–C and bottom yoke F–A–D; the central limb BA has cross-section A1 and carries coil A (N1 = 500 turns, current I1) and coil B (N2 = 300 turns, current I2); the outer limbs EF and CD have cross-section A2.]

## Short answer

Central flux 2 mWb, each outer limb 1 mWb. Reluctances: central 0.15/(μ0×1000×4e-4)=2.984e5 AT/Wb, outer 0.45/(μ0×1000×2e-4)=1.790e6 AT/Wb. Required mmf = 2e-3×2.984e5 + 1e-3×1.790e6 = 2387 AT. 500I1+300×5=2387 → I1 ≈ 1.77 A (coils aiding).

## Answer

Both coils sit on the central limb BA, so their mmfs act together on the centre limb. The centre flux divides equally between the two identical outer paths (B–C–D–A and B–E–F–A).

**Assumptions:** coils A and B aid each other; each outer path (yokes and outer limb) has area $A_2 = 2$ cm² and length BC + CD + DA = 15 + 15 + 15 = 45 cm; the central limb is 15 cm long with area $A_1 = 4$ cm²; no leakage.

```text
   E-------B-------C
   |       |       |
   |   [A: 500T]   |
   |   [B: 300T]   |
   |       | phi   |
   F-------A-------D
```

#### Fluxes

$$
\phi_c = 2\ \text{mWb}, \quad \phi_{outer} = \frac{2}{2} = 1\ \text{mWb in each outer path}
$$

#### Reluctances

$$
\begin{aligned}
S_c &= \frac{0.15}{4\pi\times10^{-7}\times1000\times4\times10^{-4}} = 2.984\times10^{5}\ \text{AT/Wb} \\
S_o &= \frac{0.45}{4\pi\times10^{-7}\times1000\times2\times10^{-4}} = 1.790\times10^{6}\ \text{AT/Wb}
\end{aligned}
$$

#### MMF required (one loop: centre limb + one outer path)

$$
\begin{aligned}
F &= \phi_c S_c + \phi_o S_o \\
&= 2\times10^{-3}\times2.984\times10^{5} + 1\times10^{-3}\times1.790\times10^{6} \\
&= 596.8 + 1790.5 = 2387.3\ \text{AT}
\end{aligned}
$$

#### Current in coil A

$$
\begin{aligned}
N_1 I_1 + N_2 I_2 &= 2387.3 \\
500\, I_1 &= 2387.3 - 300\times5 = 887.3 \\
I_1 &= 1.775\ \text{A}
\end{aligned}
$$

**Answer: I₁ ≈ 1.77 A** (coils aiding). If coil B opposes coil A, $500 I_1 = 2387.3 + 1500$ and $I_1 = 7.77$ A.

(The flux density is 5 T in every part; with real iron this would saturate, so the result holds only for the constant μr given.)

## Check yourself

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

1. How does the 2 mWb in the central limb divide?
   A. All 2 mWb goes through each outer path
   B. 1.5 mWb in one outer path and 0.5 mWb in the other
   C. Equally, 1 mWb in each outer path, since the two outer paths are identical
   D. It does not divide; it stays in the centre limb
   Answer: C. The two parallel outer paths have equal reluctance, so they share the flux equally. (Key point: Flux division)

2. Which loop does the mmf equation use?
   A. Only the centre limb: F = φc Sc
   B. Centre limb plus one outer path: F = φc Sc + φo So
   C. Both outer paths in series
   D. Centre limb plus both outer paths: F = φc Sc + 2φo So
   Answer: B. Apply mmf = Σφ S around one closed loop through the centre limb and one outer path. (Key point: Loop equation)

3. Why is So (1.79×10^6) much larger than Sc (2.98×10^5)?
   A. It has higher permeability
   B. It carries more flux
   C. The outer path is three times longer (45 cm vs 15 cm) and has half the area (2 vs 4 cm²)
   D. It has an air gap
   Answer: C. Reluctance ∝ l/A: 3 × 2 = 6 times greater. (Key point: Outer reluctance)

4. With coils aiding, what is I1 and what changes if B opposes A?
   A. I1 ≈ 4.77 A aiding; opposing needs 1.77 A
   B. I1 ≈ 7.77 A aiding; opposing needs 1.77 A
   C. I1 ≈ 1.77 A in both cases
   D. 500 I1 + 300×5 = 2387.3 gives I1 ≈ 1.77 A; opposing needs about 7.77 A
   Answer: D. Aiding: 500 I1 = 2387.3 − 1500. Opposing: 500 I1 = 2387.3 + 1500, giving 7.77 A. (Key point: Current in coil A)

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