Skip to main content

Coil current for 2 mWb in central limb of core

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).

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.]

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 A2=2A_2 = 2 cm² and length BC + CD + DA = 15 + 15 + 15 = 45 cm; the central limb is 15 cm long with area A1=4A_1 = 4 cm²; no leakage.

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

Fluxes

ϕc=2 mWb,ϕouter=22=1 mWb in each outer path\phi_c = 2\ \text{mWb}, \quad \phi_{outer} = \frac{2}{2} = 1\ \text{mWb in each outer path}

Reluctances

Sc=0.154π×10−7×1000×4×10−4=2.984×105 AT/WbSo=0.454π×10−7×1000×2×10−4=1.790×106 AT/Wb\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)

F=ϕcSc+ϕoSo=2×10−3×2.984×105+1×10−3×1.790×106=596.8+1790.5=2387.3 AT\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

N1I1+N2I2=2387.3500 I1=2387.3−300×5=887.3I1=1.775 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, 500I1=2387.3+1500500 I_1 = 2387.3 + 1500 and I1=7.77I_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.)

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electrical Machines I chapter 1 questions →