Aula 14 - Circuito RLC paralelo.ipynb

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   "source": [
    "# Circuito RLC paralelo sem fonte\n",
    "\n",
    "Jupyter Notebook desenvolvido por [Gustavo S.S.](https://github.com/GSimas)\n",
    "\n",
    "Circuitos RLC em paralelo têm diversas aplicações, como em projetos de filtros\n",
    "e redes de comunicação. Suponha que a corrente inicial I0 no indutor e a tensão inicial V0 no capacitor sejam:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large i(0) = I_0 = \\frac{1}{L} \\int_{-\\infty}^{0} v(t) dt}\n",
    "\\end{align}\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large v(0) = V_0}\n",
    "\\end{align}\n",
    "\n",
    "![](https://i.imgur.com/6gt8E4q.png)\n",
    "\n",
    "Portanto, aplicando a LKC ao nó superior fornece:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large \\frac{v}{R} + \\frac{1}{L} \\int_{-\\infty}^{t} v(\\tau) d\\tau + C \\frac{dv}{dt} = 0}\n",
    "\\end{align}\n",
    "\n",
    "Extraindo a derivada em relação a t e dividindo por C resulta em:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large \\frac{d^2v}{dt^2} + \\frac{1}{RC} \\frac{dv}{dt} + \\frac{1}{LC} v = 0}\n",
    "\\end{align}\n",
    "\n",
    "Obtemos a equação característica substituindo a primeira derivada por s e a segunda\n",
    "por s^2:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large s^2 + \\frac{1}{RC} s + \\frac{1}{LC} = 0}\n",
    "\\end{align}\n",
    "\n",
    "Assim, as raízes da equação característica são:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large s_{1,2} = -\\alpha \\pm \\sqrt{\\alpha^2 - \\omega_0^2}}\n",
    "\\end{align}\n",
    "\n",
    "onde:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large \\alpha = \\frac{1}{2RC}, \\space \\space \\space \\omega_0 = \\frac{1}{\\sqrt{LC}} }\n",
    "\\end{align}\n",
    "\n",
    "## Amortecimento Supercrítico / Superamortecimento (α > ω0)\n",
    "\n",
    "Quando α > ω0, as raízes da equação característica são reais e negativas. A resposta é:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large v(t) = A_1 e^{s_1 t} + A_2 e^{s_2 t} }\n",
    "\\end{align}\n",
    "\n",
    "## Amortecimento Crítico (α = ω0)\n",
    "\n",
    "Quando α = ω0 as raízes da equação característica são reais e iguais de modo que a resposta seja:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large v(t) = (A_1 + A_2t)e^{-\\alpha t}}\n",
    "\\end{align}\n",
    "\n",
    "## Subamortecimento (α < ω0)\n",
    "\n",
    "Quando α < ω0, nesse caso, as raízes são complexas e podem ser expressas como segue:\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large s_{1,2} = -\\alpha \\pm j\\omega_d}\n",
    "\\\\\n",
    "\\\\{\\Large \\omega_d = \\sqrt{\\omega_0^2 - \\alpha^2}}\n",
    "\\end{align}\n",
    "\n",
    "\\begin{align}\n",
    "{\\Large v(t) = e^{-\\alpha t}(A_1 cos(\\omega_d t) + A_2 sen(\\omega_d t))}\n",
    "\\end{align}\n",
    "\n",
    "![](https://i.imgur.com/3s4tDKQ.png)\n",
    "\n",
    "As constantes A1 e A2 em cada caso podem ser determinadas a partir das\n",
    "condições iniciais. Precisamos de v(0) e dv(0)/dt."
   ]
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    "**Exemplo 8.5**\n",
    "\n",
    "No circuito paralelo da Figura 8.13, determine v(t) para t > 0, supondo que v(0) = 5 V,\n",
    "i(0) = 0, L = 1 H e C = 10 mF. Considere os seguintes casos: \n",
    "\n",
    "R = 1,923 Ω, \n",
    "\n",
    "R = 5 Ω e\n",
    "\n",
    "R = 6,25 Ω .\n",
    "\n",
    "![](https://i.imgur.com/6gt8E4q.png)"
   ]
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     "text": [
      "Exemplo 8.5\n",
      "\n",
      "--------------\n",
      "\n",
      "Para R =  1.923\n",
      "Alpha: 26.001040041601662\n",
      "Omega: 10.0\n",
      "Tipo de resposta: superamortecimento\n",
      "Resposta v(t): A1*exp(-1.9999133337787*t) + A2*exp(-50.0021667494246*t)\n",
      "v(0): A1 + A2\n",
      "dv(0)/dt: -1.9999133337787*A1 - 50.0021667494246*A2\n",
      "Constante A1: -0.20855000866701326\n",
      "Constante A2: 5.208550008667014\n",
      "Resposta v(t): 5.20855000866701*exp(-50.0021667494246*t) - 0.208550008667013*exp(-1.9999133337787*t) V\n",
      "\n",
      "--------------\n",
      "\n",
      "Para R =  5\n",
      "Alpha: 10.0\n",
      "Omega: 10.0\n",
      "Tipo de resposta: amortecimento critico\n",
      "Resposta v(t): (A1 + A2*t)*exp(-10.0*t)\n",
      "v(0): A1\n",
      "dv(0)/dt: -10.0*A1 + A2\n",
      "Constante A1: 5\n",
      "Constante A2: -50.0\n",
      "Resposta v(t): (-50.0*t + 5)*exp(-10.0*t) V\n",
      "\n",
      "--------------\n",
      "\n",
      "Para R =  6.25\n",
      "Alpha: 8.0\n",
      "Omega: 10.0\n",
      "Tipo de resposta: subamortecimento\n",
      "Resposta v(t): A1*exp(-8.0*t)\n",
      "v(0): A1\n",
      "dv(0)/dt: -8.0*A1\n",
      "Constante A1: 5\n",
      "Constante A2: -6.666666666666667\n",
      "Resposta v(t): 5*exp(-8.0*t) V\n"
     ]
    }
   ],
   "source": [
    "print(\"Exemplo 8.5\")\n",
    "\n",
    "from sympy import *\n",
    "\n",
    "m = 10**(-3) #definicao de mili\n",
    "L = 1\n",
    "C = 10*m\n",
    "v0 = 5\n",
    "i0 = 0\n",
    "\n",
    "A1 = symbols('A1')\n",
    "A2 = symbols('A2')\n",
    "t = symbols('t')\n",
    "\n",
    "def sqrt(x, root = 2): #definir funcao para raiz\n",
    "    y = x**(1/root)\n",
    "    return y\n",
    "\n",
    "print(\"\\n--------------\\n\")\n",
    "\n",
    "## PARA R = 1.923\n",
    "R = 1.923\n",
    "\n",
    "print(\"Para R = \", R)\n",
    "\n",
    "def resolve_rlc(R,L,C):\n",
    "\n",
    "    alpha = 1/(2*R*C)\n",
    "    omega = 1/sqrt(L*C)\n",
    "    print(\"Alpha:\",alpha)\n",
    "    print(\"Omega:\",omega)\n",
    "    \n",
    "    s1 = -alpha + sqrt(alpha**2 - omega**2)\n",
    "    s2 = -alpha - sqrt(alpha**2 - omega**2)\n",
    "\n",
    "    def rlc(alpha,omega): #funcao para verificar tipo de amortecimento\n",
    "        resposta = \"\"\n",
    "        if alpha > omega:\n",
    "            resposta = \"superamortecimento\"\n",
    "            v = A1*exp(s1*t) + A2*exp(s2*t)\n",
    "        elif alpha == omega:\n",
    "            resposta = \"amortecimento critico\"\n",
    "            v = (A1 + A2*t)*exp(-alpha*t)\n",
    "        else:\n",
    "            resposta = \"subamortecimento\"\n",
    "            v = exp(-alpha*t)*(A1*cos(omega_d*t) + A2*sin(omega_d*t))\n",
    "        return resposta,v\n",
    "\n",
    "    resposta,v = rlc(alpha,omega)\n",
    "    print(\"Tipo de resposta:\",resposta)\n",
    "    print(\"Resposta v(t):\",v)\n",
    "    print(\"v(0):\",v.subs(t,0))\n",
    "    print(\"dv(0)/dt:\",v.diff(t).subs(t,0))\n",
    "    \n",
    "    return alpha,omega,s1,s2,resposta,v\n",
    "\n",
    "alpha,omega,s1,s2,resposta,v = resolve_rlc(R,L,C)\n",
    "\n",
    "#v(0) = 5 = A1 + A2 -> A2 = 5 - A1\n",
    "#dv(0)/dt = -2A1 - 50A2\n",
    "#C*dv(0)/dt + i(0) + v(0)/R = 0\n",
    "    #0.01*(-2A1 - 50A2) + 0 + 5/1.923 = 0\n",
    "    #(-2A1 -50(5 - A1)) = -5/(1.923*0.01)\n",
    "    #48A1 = 250 - 5/(1.923*0.01)\n",
    "A1 = (250 - 5/(1.923*0.01))/48\n",
    "print(\"Constante A1:\",A1)\n",
    "A2 = 5 - A1\n",
    "print(\"Constante A2:\",A2)\n",
    "v = A1*exp(s1*t) + A2*exp(s2*t)\n",
    "print(\"Resposta v(t):\",v,\"V\")\n",
    "\n",
    "print(\"\\n--------------\\n\")\n",
    "\n",
    "## PARA R = 5\n",
    "R = 5\n",
    "\n",
    "A1 = symbols('A1')\n",
    "A2 = symbols('A2')\n",
    "\n",
    "print(\"Para R = \", R)\n",
    "\n",
    "alpha,omega,s1,s2,resposta,v = resolve_rlc(R,L,C)\n",
    "\n",
    "#v(t) = (A1 + A2t)e^(-alpha*t)\n",
    "\n",
    "#v(0) = A1 = 5\n",
    "A1 = 5\n",
    "#C*dv(0)/dt + i(0) + v(0)/R = 0\n",
    "    #0.01(-10A1 + A2) + 0 + 5/5 = 0\n",
    "    #0.01A2 = -1 + 0.5\n",
    "A2 = (-1 + 0.5)/0.01\n",
    "\n",
    "print(\"Constante A1:\",A1)\n",
    "print(\"Constante A2:\",A2)\n",
    "v = (A1 + A2*t)*exp(-alpha*t)\n",
    "print(\"Resposta v(t):\",v,\"V\")\n",
    "\n",
    "print(\"\\n--------------\\n\")\n",
    "\n",
    "\n",
    "## PARA R = 6.25\n",
    "R = 6.25\n",
    "\n",
    "A1 = symbols('A1')\n",
    "A2 = symbols('A2')\n",
    "\n",
    "print(\"Para R = \", R)\n",
    "\n",
    "omega_d = sqrt(omega**2 - alpha**2)\n",
    "alpha,omega,s1,s2,resposta,v = resolve_rlc(R,L,C)\n",
    "\n",
    "#v(t) = e^-(alpha*t)*(A1cos(wd*t) + A2sen(wd*t))\n",
    "\n",
    "#v(0) = A1 = 5\n",
    "A1 = 5\n",
    "#C*dv(0)/dt + i(0) + v(0)/R = 0\n",
    "    #0.01*(-8A1 + 6A2) + 0 + 5/6.25 = 0\n",
    "    #-0.4 + 0.06A2 = -5/6.25\n",
    "A2 = (-5/6.25 + 0.4)/0.06\n",
    "\n",
    "print(\"Constante A1:\",A1)\n",
    "print(\"Constante A2:\",A2)\n",
    "v = exp(-alpha*t)*(A1*cos(omega_d*t) + A2*sin(omega_d*t))\n",
    "print(\"Resposta v(t):\",v,\"V\")"
   ]
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   "source": [
    "**Problema Prático 8.5**\n",
    "\n",
    "Na Figura 8.13, seja R = 2 Ω, L = 0,4 H, C = 25 mF, v(0) = 0, e i(0) = 50 mA. Determine v(t) para t > 0.\n",
    "\n",
    "![](https://i.imgur.com/6gt8E4q.png)"
   ]
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     "text": [
      "Problema Prático 8.5\n",
      "Alpha: 10.0\n",
      "Omega: 10.0\n",
      "Tipo de resposta: amortecimento critico\n",
      "Resposta v(t): (A1 + A2*t)*exp(-10.0*t)\n",
      "v(0): A1\n",
      "dv(0)/dt: -10.0*A1 + A2\n",
      "Constante A1: 0\n",
      "Constante A2: -2.0\n",
      "Resposta v(t): -2.0*t*exp(-10.0*t) V\n"
     ]
    }
   ],
   "source": [
    "print(\"Problema Prático 8.5\")\n",
    "\n",
    "R = 2\n",
    "L = 0.4\n",
    "C = 25*m\n",
    "v0 = 0\n",
    "i0 = 50*m\n",
    "\n",
    "A1 = symbols('A1')\n",
    "A2 = symbols('A2')\n",
    "\n",
    "alpha,omega,s1,s2,resposta,v = resolve_rlc(R,L,C)\n",
    "\n",
    "#C*dv(0)/dt + i(0) + v(0)/R = 0\n",
    "    #C*(-10A1 + A2) + i0 + v(0)/2 = 0\n",
    "    #v(0) = 0 = A1\n",
    "    #C*A2 = -i0\n",
    "A2 = -i0/C\n",
    "A1 = 0\n",
    "\n",
    "print(\"Constante A1:\",A1)\n",
    "print(\"Constante A2:\",A2)\n",
    "v = (A1 + A2*t)*exp(-10.0*t)\n",
    "print(\"Resposta v(t):\",v,\"V\")"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exemplo 8.6**\n",
    "\n",
    "Determine v(t) para t > 0 no circuito RLC da Figura 8.15.\n",
    "\n",
    "![](https://i.imgur.com/07Ueco1.png)"
   ]
  },
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   "execution_count": 12,
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     "text": [
      "Exemplo 8.6\n",
      "V0: 25.0 V\n",
      "i0: -0.5 A\n",
      "Alpha: 500.0\n",
      "Omega: 353.5533905932738\n",
      "Tipo de resposta: superamortecimento\n",
      "Resposta v(t): A1*exp(-146.446609406726*t) + A2*exp(-853.553390593274*t)\n",
      "v(0): A1 + A2\n",
      "dv(0)/dt: -146.446609406726*A1 - 853.553390593274*A2\n",
      "Constante A1: 30.15536723163842\n",
      "Constante A2: -5.155367231638418\n",
      "Resposta v(t): -5.15536723163842*exp(-853.553390593274*t) + 30.1553672316384*exp(-146.446609406726*t) V\n"
     ]
    }
   ],
   "source": [
    "print(\"Exemplo 8.6\")\n",
    "\n",
    "u = 10**(-6) #definicao de micro\n",
    "Vs = 40\n",
    "L = 0.4\n",
    "C = 20*u\n",
    "\n",
    "A1 = symbols('A1')\n",
    "A2 = symbols('A2')\n",
    "\n",
    "#Para t < 0\n",
    "v0 = Vs*50/(50 + 30)\n",
    "i0 = -Vs/(50 + 30)\n",
    "\n",
    "print(\"V0:\",v0,\"V\")\n",
    "print(\"i0:\",i0,\"A\")\n",
    "\n",
    "#Para t > 0\n",
    "#C*dv(0)/dt + i(0) + v(0)/50 = 0\n",
    "    #20u*dv(0)/dt - 0.5 + 0.5 = 0\n",
    "    #dv(0)/dt = 0\n",
    "\n",
    "R = 50\n",
    "alpha,omega,s1,s2,resposta,v = resolve_rlc(R,L,C)\n",
    "\n",
    "#v(0) = 25 = A1 + A2\n",
    "    #A1 = 25 - A2\n",
    "#dv(0)/dt = -146A1 - 854A2 = 0\n",
    "    #-146(25 - A2) - 854A2 = 0\n",
    "    #146A2 - 854A2 = 3650\n",
    "    #-708A2 = 3650\n",
    "A2 = -3650/708\n",
    "A1 = 25 - A2\n",
    "\n",
    "print(\"Constante A1:\",A1)\n",
    "print(\"Constante A2:\",A2)\n",
    "\n",
    "v = A1*exp(s1*t) + A2*exp(s2*t)\n",
    "print(\"Resposta v(t):\",v,\"V\")"
   ]
  }
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