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A three-legged core has a center limb with a coil. Two outer limbs are parallel paths. Center limb length l_c = 0.2 m , outer limb lengths l_o = 0.3 m each, same A = 600 mm² = 6×10⁻⁴ m² . N = 300 , I = 3 A , μᵣ = 2000 for iron. Find total flux and flux in each outer limb.
1Rp=1Rl+1Rl⟹Rp=Rl2=133,333.32=66,666.6 Av/Wbthe fraction with numerator 1 and denominator script cap R sub p end-fraction equals the fraction with numerator 1 and denominator script cap R sub l end-fraction plus the fraction with numerator 1 and denominator script cap R sub l end-fraction ⟹ script cap R sub p equals the fraction with numerator script cap R sub l and denominator 2 end-fraction equals the fraction with numerator 133 comma 333.3 and denominator 2 end-fraction equals 66 comma 666.6 Av/Wb
Última actualización: Febrero 2026
Debido a la perfecta simetría del circuito magnético: circuitos magneticos ejercicios resueltos
Thus: [ ℛ = \frac0.32.513 × 10^-7 \approx 1.194 × 10^6 \text A·t/Wb ]
Rg=lgμ0⋅Ascript cap R sub g equals the fraction with numerator l sub g and denominator mu sub 0 center dot cap A end-fraction
Las longitudes deben estar siempre en metros ( ) y las áreas en metros cuadrados ( m2m squared A three-legged core has a center limb with a coil
: Visual tutorials that explain how to apply formulas to physical diagrams like toroids and transformers.
"Compute reluctance using $\mathcalR = \fracl\mu A$ with $\mu = \mu_r \mu_0$ for the entire core, including the gap."
I=FN=497.41500=0.995 A≈1 Acap I equals the fraction with numerator cap F and denominator cap N end-fraction equals 497.41 over 500 end-fraction equals 0.995 A is approximately equal to 1 A N = 300 , I = 3 A , μᵣ = 2000 for iron
El flujo suele "dispersarse" en los bordes del entrehierro (efecto de borde), pero en ejercicios básicos se asume que el área es la misma que la del núcleo.
: Ejercicios desde mallas sencillas hasta circuitos complejos con resolución gráfica. Repositorio de ResearchGate
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