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El circuito magneticode la figura, tiene materialesde permeabilidadesμ,2μ, 4μ. Si el radio medioes
“r”.Halle la femque debe aplicar para que circule un flujoΦ
Debemoshallarlafmm enfunciónde Φ, r y las permeabilidadesμ,2μ, 4μ, pero primerohallamosel
circuitoeléctricoequivalente.
Hallamoslareluctancias,segúnlaecuacion:
𝑅 =
𝑙𝑚
𝜇 ∗ 𝐴
𝑅1 =
𝜋 ∗ 𝑟
2 ∗ 𝜇 ∗ 𝐴
; 𝑅2 =
𝜋 ∗ 𝑟
2 ∗ (2 ∗ 𝜇) ∗ 𝐴
=
𝜋 ∗ 𝑟
4 ∗ 𝜇 ∗ 𝐴
; 𝑅3 =
𝜋 ∗ 𝑟
(4 ∗ 𝜇) ∗ 𝐴
=
𝜋 ∗ 𝑟
4 ∗ 𝜇 ∗ 𝐴
Hallamoslareluctanciaequivalente:
𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3 =
𝜋 ∗ 𝑟
2 ∗ 𝜇 ∗ 𝐴
+
𝜋 ∗ 𝑟
4 ∗ 𝜇 ∗ 𝐴
+
𝜋 ∗ 𝑟
4 ∗ 𝜇 ∗ 𝐴
𝑅𝑒𝑞 =
𝜋 ∗ 𝑟
𝜇 ∗ 𝐴
Hallamoslafmmpor la que circulaun flujo Φ utilizandolaleyde Hopkinson:
𝑓
𝑚𝑚 = 𝛷 ∗ 𝑅𝑒𝑞 = 𝛷 ∗
𝜋 ∗ 𝑟
𝜇 ∗ 𝐴
=
𝛷 ∗ 𝜋 ∗ 𝑟
𝜇 ∗ 𝐴
En la figura,se muestra un circulomagnetico. Encuentre a) La reactancia total del circuito. b) Halle el
flujoneto del circuito
Siendoel circuitoeléctricoequivalente
Hallamoslareluctanciasyfuerza magnetomotricesdelcircuitoequivalente
𝑅1 =
𝜋 ∗ 𝑟
𝜇 ∗ 𝐴
; 𝑅2 =
𝜋 ∗ 𝑟
2 ∗ 𝜇 ∗ 𝐴
𝑓𝑚𝑚1 = 𝑁𝑖 ; 𝑓𝑚𝑚2 = 2𝑁𝑖
Comolas fuerzasmagnetomotricesestánenserie yestánenpolaridadcontraria,estasse restaran:
𝑓𝑚𝑚 = 𝑓𝑚𝑚2 − 𝑓𝑚𝑚1 = 2 ∗ 𝑁𝑖 − 𝑁𝑖 = 𝑁𝑖
a) la reluctanciatotal del circuitoes“Req” siendo:
𝑅𝑒𝑞 = 𝑅1 + 𝑅2 =
𝜋 ∗ 𝑟
𝜇 ∗ 𝐴
+
𝜋 ∗ 𝑟
2 ∗ 𝜇 ∗ 𝐴
=
3 ∗ 𝜋 ∗ 𝑟
2 ∗ 𝜇 ∗ 𝐴
b) Para hallarel flujonetose usalaleyde Hopkinson:
𝑓
𝑚𝑚 = 𝛷 ∗ 𝑅𝑒𝑞 → 𝛷 =
𝑓
𝑚𝑚
𝑅𝑒𝑞
=
𝑁 ∗ 𝑖
3 ∗ 𝜋 ∗ 𝑟
2 ∗ 𝜇 ∗ 𝐴
=
2 ∗ 𝜇 ∗ 𝐴 ∗ 𝑁 ∗ 𝑖
3 ∗ 𝜋 ∗ 𝑟

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Fis 200

  • 1. El circuito magneticode la figura, tiene materialesde permeabilidadesμ,2μ, 4μ. Si el radio medioes “r”.Halle la femque debe aplicar para que circule un flujoΦ Debemoshallarlafmm enfunciónde Φ, r y las permeabilidadesμ,2μ, 4μ, pero primerohallamosel circuitoeléctricoequivalente. Hallamoslareluctancias,segúnlaecuacion: 𝑅 = 𝑙𝑚 𝜇 ∗ 𝐴 𝑅1 = 𝜋 ∗ 𝑟 2 ∗ 𝜇 ∗ 𝐴 ; 𝑅2 = 𝜋 ∗ 𝑟 2 ∗ (2 ∗ 𝜇) ∗ 𝐴 = 𝜋 ∗ 𝑟 4 ∗ 𝜇 ∗ 𝐴 ; 𝑅3 = 𝜋 ∗ 𝑟 (4 ∗ 𝜇) ∗ 𝐴 = 𝜋 ∗ 𝑟 4 ∗ 𝜇 ∗ 𝐴
  • 2. Hallamoslareluctanciaequivalente: 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3 = 𝜋 ∗ 𝑟 2 ∗ 𝜇 ∗ 𝐴 + 𝜋 ∗ 𝑟 4 ∗ 𝜇 ∗ 𝐴 + 𝜋 ∗ 𝑟 4 ∗ 𝜇 ∗ 𝐴 𝑅𝑒𝑞 = 𝜋 ∗ 𝑟 𝜇 ∗ 𝐴 Hallamoslafmmpor la que circulaun flujo Φ utilizandolaleyde Hopkinson: 𝑓 𝑚𝑚 = 𝛷 ∗ 𝑅𝑒𝑞 = 𝛷 ∗ 𝜋 ∗ 𝑟 𝜇 ∗ 𝐴 = 𝛷 ∗ 𝜋 ∗ 𝑟 𝜇 ∗ 𝐴
  • 3. En la figura,se muestra un circulomagnetico. Encuentre a) La reactancia total del circuito. b) Halle el flujoneto del circuito Siendoel circuitoeléctricoequivalente Hallamoslareluctanciasyfuerza magnetomotricesdelcircuitoequivalente 𝑅1 = 𝜋 ∗ 𝑟 𝜇 ∗ 𝐴 ; 𝑅2 = 𝜋 ∗ 𝑟 2 ∗ 𝜇 ∗ 𝐴 𝑓𝑚𝑚1 = 𝑁𝑖 ; 𝑓𝑚𝑚2 = 2𝑁𝑖 Comolas fuerzasmagnetomotricesestánenserie yestánenpolaridadcontraria,estasse restaran: 𝑓𝑚𝑚 = 𝑓𝑚𝑚2 − 𝑓𝑚𝑚1 = 2 ∗ 𝑁𝑖 − 𝑁𝑖 = 𝑁𝑖
  • 4. a) la reluctanciatotal del circuitoes“Req” siendo: 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 = 𝜋 ∗ 𝑟 𝜇 ∗ 𝐴 + 𝜋 ∗ 𝑟 2 ∗ 𝜇 ∗ 𝐴 = 3 ∗ 𝜋 ∗ 𝑟 2 ∗ 𝜇 ∗ 𝐴 b) Para hallarel flujonetose usalaleyde Hopkinson: 𝑓 𝑚𝑚 = 𝛷 ∗ 𝑅𝑒𝑞 → 𝛷 = 𝑓 𝑚𝑚 𝑅𝑒𝑞 = 𝑁 ∗ 𝑖 3 ∗ 𝜋 ∗ 𝑟 2 ∗ 𝜇 ∗ 𝐴 = 2 ∗ 𝜇 ∗ 𝐴 ∗ 𝑁 ∗ 𝑖 3 ∗ 𝜋 ∗ 𝑟