Cc.x Circuits exercise - JulTob/Mathematics GitHub Wiki
El circuito siguiente representa una fuente de corriente controlada por tensión. Obtenga el valor de la corriente Io en función de la tensión V1 y de las resistencias. ¿Depende la corriente del valor de la carga RL?
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V1@{ shape: circle, label: "V1🔋" }
style V1 fill:#080,stroke:#050,stroke-width:4px
R2@{ shape: rect, label: "2·R 💡💡" }
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V1 o--o|Ⓐ| R2
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R2 o--o N1
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R01@{ shape: rect, label: "R💡" }
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N1 o--o R01
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N1 o--o|"－"| OP
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R01 o--o|Ⓒ| N2
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OP o--o N2
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GR@{ shape: trap-t, label: "Ground🪫" }
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R20@{ shape: rect, label: "2·R 💡💡" }
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R20 o--o|Ⓕ| GR
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R02 o--o|Ⓓ| N2
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N3 o--o R20
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N3 o--o|"＋"| OP
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N3 o--o R02
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V2@{ shape: circle, label: "V0📡" }
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N2 o--o V2
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RL@{ shape: lin-rect, label: "RL 🔌" }
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N3 o----o RL
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GR2@{ shape: trap-t, label: "Gr🪫" }
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RL o--o|Ⓖ| GR2
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%% Force straight lines
linkStyle default interpolate basis;
-
Con realimentación negativa
$>> ccv$ $v+ = v- = V_x$ - i+ = i- = 0
-
$① 🟠IⒶⒷ = 🟡IⒷⒸ$ -
$② 🟡IⒹⒺ = 🟠IⒺⒻ + 🔵IⒺⒼ$ -
$③ 🔵IⒺⒼ = I₀ = \frac{V_x}{R_L}$ -
$①→④ 🟠\frac{(V₁-V_x)}{2R} = 🟡\frac{(V_x-V₀)}{R}$ $④ V₁+2·V₀ = 3·V_x$
-
$②→⑤ 🟡\frac{(V₀-V_x)}{R} = 🟠\frac{V_x}{2R} + 🔵\frac{V_x}{R_L}$ - ⑤
$V₀ = V_x + (\frac{V_x}{2}) + V_x(\frac{R}{R_L})$
- ⑤
-
$④＆⑤→⑥ V₁ + 2·(V_x + (V_x:2) + V_x(R:R_L)) = 3·V_x$ $⑥ V₁ = V_x ( 3 - 2 - 1 - (2R:R_L) )$ $⑥ V₁ = V_x ( -2R : R_L )$
-
$③ 🔵 I₀ = V₁(R_L:(-2R)):R_L = V₁(1:(-2R)) = - V₁:2R$
- No depende de RL
- Fuente de corriente independiente de la resistencia de carga
Plantee las ecuaciones de malla en términos de las corrientes de malla I1, I2 en el siguiente circuito. Especifique el valor de las tensiones en las bobinas en función de las mismas corrientes I1, I2. DATOS: Vs= 180 ∟0° (V); M=4H; L1 =5H; L2 =6H; R=10 Ω; C=1mF; 𝜔=20rad/s
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L2@{ shape: cyl, label: "💈L2" }
style L2 fill:#604,stroke:#E09,stroke-width:4px
N1@{ shape: circ, label: "🔆" }
style N1 fill:#000,stroke:#444,stroke-width:2px
N2@{ shape: circ, label: "🔆" }
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C1@{ shape: procs, label: "C1 ⏳" }
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R@{ shape: rect, label: "R💡" }
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A[Battery 🔋∿Vs🔋] --o |<span style='color:orange;'>I1+I2</span>| L1
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L1 o--o N1
L1 o-.-o |<span style='color:pink;'>🧲M🧲</span>| L2
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N1 o--o |<span style='color:yellow;'>I2</span>|R
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N1 o--o |<span style='color:red;'>I1</span>|L2
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L2 o--o C1
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N2 o--o C1
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A o--o N2
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N2 o--o R
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class A source;
-
$VL_1 = 𝚒·𝜔·L_1·𝕀₁ - 𝚒·𝜔·M·𝕀₂$ $= 𝚒·100·𝕀₁ - 𝚒·80·𝕀₂$
-
$VL_2 = 𝚒·𝜔·L_2·𝕀₂ - 𝚒·𝜔·M·𝕀₁$ $= 𝚒·120·𝕀₂ - 𝚒·80·𝕀₁$
- Malla 1:
$-V_s + V_{L_1} + V_R = 0$ $-V_s + (𝚒·100·𝕀₁ - 𝚒·80·𝕀₂) + (R(𝕀₁ - 𝕀₂)) = 0$ - 🟢
$180 = (𝚒·100+10)𝕀₁ + (-10 - 𝚒·80)𝕀₂$
-
$Malla 2: -R(𝕀₁-𝕀₂) + V_{L_2} + V_c = 0$ $-10(𝕀₁-𝕀₂) + (𝚒·120·𝕀₂ - 𝚒·80·𝕀₁) + 𝕀₂·Z_c = 0$ $-10𝕀₁ + 10𝕀₂ + 𝚒·120·𝕀₂ - 𝚒·80·𝕀₁ - 50·𝕀₂ = 0$ - 🟢
$(-10 - 𝚒·80)𝕀₁ + (10+70𝚒)𝕀₂ =0$
El circuito mostrado ha estado durante mucho tiempo con el interruptor en la posición A. En el instante t=0, el interruptor pasa a la posición B. Calcular como evoluciona la tensión en bornes del condensador Vc(t) para t ≥0.
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Sw@{ shape: diamond, label: "Switch 🕹️" }
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NA@{ shape: circ, label: "🔆" }
R3@{ shape: stadium, label: "💡R" }
C@{ shape: processes, label: "C(Vc)" }
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Vs o---o R1
R1 o----o Sw
Sw o-.-o|t<0| R2
Sw o-.-o|t>0| NA
R2 o---o NA
NA o---o R3
R3 o---o Gr
NA o---o C
C o---o Gr
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- si
$t⟶∞$ :- C es un abierto
$V_c(∞)= I·R/2 = (12:R)(R:2)= 6[V]$
$s = -1:(Req·C)= -1:5k·Ⅹ(-6) = -Ⅹ(3):5 = -200 [s̈]$
