MaxPowerTransfer.ipynb

{
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   "metadata": {},
   "source": [
    "# Maximum Power Transfer Theorem\n",
    "Consider a source-load single loop circuit, as presented in Fig. 2.5.11a and the following\n",
    "problem:\n",
    "For the given source, characterized by parameters $E_\\mathrm{o}$ and $R_\\mathrm{t}$, find the load resistance ($R_\\mathrm{l}$) such\n",
    "that the maximum power available from the source is transferred to this load."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a38f2982",
   "metadata": {},
   "source": [
    "The power absorbed by the load is:  \n",
    "\n",
    "\\begin{equation}\\tag{1}\n",
    "    P_\\mathrm{l} = I^2R_\\mathrm{l} = \\left( \\frac{E_\\mathrm{o}}{R_\\mathrm{t} + R_\\mathrm{l}}\\right)^2R_\\mathrm{l} = f(R_\\mathrm{l})\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "887601d5",
   "metadata": {},
   "source": [
    "To find the value of $R_\\mathrm{l}$ that maximizes this power, the differential calculus to find where the\n",
    "derivative equals zero can be used:  \n",
    "\n",
    "\\begin{equation}\\tag{2}\n",
    "    \\frac{\\mathrm{d}P_\\mathrm{l}}{\\mathrm{d}R_\\mathrm{l}} = 0 \\implies R_\\mathrm{l} = R_\\mathrm{t}\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "4edd5c1c",
   "metadata": {},
   "source": [
    "Or, to put it another way, $f'(R_\\mathrm{t}) = 0$."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "05a8004c",
   "metadata": {},
   "source": [
    "The proof for the above relationship:  \n",
    "\n",
    "$$ f'(R_\\mathrm{l}) = I^2 - 2R_\\mathrm{l}\\,\\frac{E_\\mathrm{o}^2}{(R_\\mathrm{t} + R_\\mathrm{l})^3}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "04caeef9",
   "metadata": {},
   "source": [
    "$$ f'(R_\\mathrm{t}) = \\frac{E_\\mathrm{o}^2}{4R_\\mathrm{t}^2} - 2R_\\mathrm{t}\\,\\frac{E_\\mathrm{o}^2}{8R_\\mathrm{t}^3} = 0$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "0b865851",
   "metadata": {},
   "source": [
    "The maximum power delivered by a source represented by its Thevenin equivalent is attained\n",
    "when the load resistance ($R_\\mathrm{l}$) is equal to the internal resistance ($R_\\mathrm{t}$)."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "92d23712",
   "metadata": {},
   "source": [
    "In general, the roots ($R$) of $f'(R_\\mathrm{l})$ are given by:  \n",
    "\n",
    "$$ f'(R) = 0$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "5b707a77",
   "metadata": {},
   "source": [
    "$$ E_\\mathrm{o}^2 - 2R\\,\\frac{E_\\mathrm{o}^2}{R_\\mathrm{t} + R} = 0$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "152b6db6",
   "metadata": {},
   "source": [
    "$$ 1 - \\frac{2R}{R_\\mathrm{t}+R} = 0$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ff62355b",
   "metadata": {},
   "source": [
    "$$ \\frac{2R}{R_\\mathrm{t}+R} = 1$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c64e7bbf",
   "metadata": {},
   "source": [
    "$$ R_\\mathrm{t} + R = 2R$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3d8c12f2",
   "metadata": {},
   "source": [
    "$$ R = R_\\mathrm{t}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "5a388c83",
   "metadata": {},
   "source": [
    "The normalized plot $P_\\mathrm{l}\\,\\mathrm{/}\\,P_\\mathrm{l}^\\max$ as a function of $R_\\mathrm{l}\\,/\\,R_\\mathrm{t}$  is presented in Fig. 2.6.1, together with the system\n",
    "efficiency plot."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "9dc25410",
   "metadata": {},
   "source": [
    "**Efficiency of power transfer** is defined as ratio of the power delivered to a load from a\n",
    "practical source to the power supplied by the emf (the Thevenin equivalent voltage or the Norton equivalent current).  \n",
    "\n",
    "\\begin{equation}\\tag{3}\n",
    "    \\eta \\% = \\frac{P_\\mathrm{l}}{P_{E_\\mathrm{o}}}\\,100\\% = \\frac{I^2R_\\mathrm{l}}{IE_\\mathrm{o}}\\,100\\% = \\frac{R_\\mathrm{l}}{R_\\mathrm{t}+R_\\mathrm{l}}\\,100\\%\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e5d6cefc",
   "metadata": {},
   "source": [
    "The above follows from the following:  Via KVL and Ohm's law,  \n",
    "\n",
    "$$ U - R_\\mathrm{l}I = E_\\mathrm{o} - R_\\mathrm{t}I - R_\\mathrm{l}I= 0$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f8869923",
   "metadata": {},
   "source": [
    "$$ E_\\mathrm{o} = I(R_\\mathrm{t} + R_\\mathrm{l})$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3c04eb6a",
   "metadata": {},
   "source": [
    "$$ \\frac{1}{R_\\mathrm{t} + R_\\mathrm{l}} = \\frac{I}{E_\\mathrm{o}}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "41822214",
   "metadata": {},
   "source": [
    "Under **maximum power transfer** conditions:"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "92f7f266",
   "metadata": {},
   "source": [
    "\\begin{equation}\\tag{4a}\n",
    "    P_\\mathrm{l} = P_\\mathrm{l}^\\max = f(R_\\mathrm{t}) = \\frac{E_\\mathrm{o}^2}{4 R_\\mathrm{t}}\n",
    "\\end{equation}"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "5dc22994",
   "metadata": {},
   "source": [
    "\\begin{equation}\\tag{4b}\n",
    "    I = \\frac{E_\\mathrm{o}}{2R_\\mathrm{t}}\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "968bd6a3",
   "metadata": {},
   "source": [
    "and,\n",
    "\\begin{equation}\\tag{4c}\n",
    "    \\eta \\% = 50\\%\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "92ade5f9",
   "metadata": {},
   "source": [
    " $$\\frac{f(R_\\mathrm{l})}{f(R_\\mathrm{t})} = \\frac{4R_\\mathrm{l}R_\\mathrm{t}}{(R_\\mathrm{t}+R_\\mathrm{l})^2}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "45339836",
   "metadata": {},
   "source": [
    "### Example 2.6.2\n",
    "In the circuit shown in Fig. 2.6.2, suppose $E_x$ is adjustable. Find its value such that the\n",
    "maximum power is transferred from A to B. What is the value of $P_\\mathrm{l}^\\max$.  \n",
    "\n",
    "![Fig-2.6.2](Figures/Fig-2.6.2.png)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e8354cec",
   "metadata": {},
   "source": [
    "There are two ways to find the Norton/Thevenin equivalents (in terms of $E_x$) of subcircuit B."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "87893574",
   "metadata": {},
   "source": [
    "#### Method 1\n",
    "\n",
    "Via KVL and Ohm's Law,  \n",
    "\n",
    "$$ 200\\Omega\\,I + 50\\Omega\\,I - E_x = 0$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "96c2a661",
   "metadata": {},
   "source": [
    "$$ 250\\Omega\\,I = E_x $$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3b11fea6",
   "metadata": {},
   "source": [
    "$$E_\\mathrm{o} = 200\\Omega\\,I = \\frac{2}{2.5}\\,E_x$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "5c58a00a",
   "metadata": {},
   "source": [
    "$$ G_\\mathrm{t} = \\frac{1}{200\\Omega} + \\frac{1}{50\\Omega} = \\frac{5}{200\\Omega} = \\frac{1}{40}\\,\\mathrm{S}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e0805f45",
   "metadata": {},
   "source": [
    "$$ R_\\mathrm{t} = 40\\Omega$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "cf464e52",
   "metadata": {},
   "source": [
    "$$ J_\\mathrm{s} = E_\\mathrm{o}G_\\mathrm{t} = \\frac{1}{50\\Omega}\\,E_\\mathrm{x}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ba5793c3",
   "metadata": {},
   "source": [
    "#### Method 2\n",
    "\n",
    "The second method involves obtaining the Norton equivalent of the subcircuit consisting of the $50\\Omega$ resistance and the voltage source $E_x$ first, and then combining that subcircuit with the $200\\Omega$ resistance that is in parallel with said subcircuit into one subcircuit.  The internal conductance ($G_\\mathrm{t}$) of that subcircuit is equal to the sum of the conductances that are in parallel with each other therein."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f4ef67de",
   "metadata": {},
   "source": [
    "The power absorbed by the load ($P_\\mathrm{l}$) is:  \n",
    "\n",
    "\\begin{align*}\n",
    "    P_\\mathrm{l} = IU &= -U(J_\\mathrm{s} - G_\\mathrm{t}U) \\\\\n",
    "     &= UG_\\mathrm{t}(U - E_\\mathrm{o})\n",
    "\\end{align*}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "23abc4f6",
   "metadata": {},
   "source": [
    "$$ \\frac{\\mathrm{d}P_\\mathrm{l}}{\\mathrm{d}U} = G_\\mathrm{t}(2U-E_\\mathrm{o})$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3511c786",
   "metadata": {},
   "source": [
    "We need to find $E_x$ such that:  \n",
    "\n",
    "\\begin{equation}\n",
    "    0 = 2U-E_\\mathrm{o} \n",
    "\\end{equation}     "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "85cd94c9",
   "metadata": {},
   "source": [
    "$$ E_\\mathrm{o}= 20\\mathrm{V} = 2U$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "09da3ca3",
   "metadata": {},
   "source": [
    "Thus,  \n",
    "\n",
    "\\begin{equation}\\tag{Ex-2.1}\n",
    "    U = 10\\mathrm{V}\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b459f1ae",
   "metadata": {},
   "source": [
    "Via equation (8) of the [Thevenin-Norton](Thevenin-Norton.ipynb#Active-Two-Terminal-Subcircuit) section,  \n",
    "\n",
    "$$ U = 10\\mathrm{V} = 20\\mathrm{V} - 100\\Omega\\,I$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "45af7f17",
   "metadata": {},
   "source": [
    "$$ 100\\Omega\\,I = 10\\mathrm{V}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "5adee948",
   "metadata": {},
   "source": [
    "\\begin{equation}\\tag{Ex-2.2}\n",
    "    I = 0.1\\mathrm{A}\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "eaf184f7",
   "metadata": {},
   "source": [
    "Via equation (Ex-2.1), KVL, and Ohm's Law,  \n",
    "\n",
    "\\begin{equation}\\tag{Ex-2.3}\n",
    "    U = 10\\mathrm{V} = 200\\Omega\\,I_1\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1f33feef",
   "metadata": {},
   "source": [
    "$$ I_1 = \\frac{1}{20}\\,\\mathrm{A} = 50\\mathrm{mA}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e0e21661",
   "metadata": {},
   "source": [
    "Via KCL and equation (Ex-2.2),  \n",
    "\n",
    "\\begin{equation}\\tag{Ex-2.4}\n",
    "    I = 0.1\\mathrm{A} = I_1 + I_2\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "16166aeb",
   "metadata": {},
   "source": [
    "Via the above two equations and substitution,  \n",
    "\n",
    "\\begin{equation}\\tag{Ex-2.5}\n",
    "    I_2 = 100\\mathrm{mA} - 50\\mathrm{mA} = 50\\mathrm{mA}\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "9b5f0f2b",
   "metadata": {},
   "source": [
    "Finally, via KVL and Ohm's Law,  \n",
    "\n",
    "$$ 200\\Omega\\,I_1 - 50\\Omega\\,I_2 - E_x = 0$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "8ddd8f91",
   "metadata": {},
   "source": [
    "\\begin{equation}\\tag{Ex-2.6}\n",
    "    E_x = 200\\Omega\\,I_1 - 50\\Omega\\,I_2 = 150\\Omega\\cdot 50\\mathrm{mA} = 7.5\\mathrm{V}\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "id": "895d7b20",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "7.5"
      ]
     },
     "execution_count": 1,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "E_x = 150*0.05\n",
    "E_x"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "8c06ad73",
   "metadata": {},
   "source": [
    "## Drill Problems"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "8c41bac5",
   "metadata": {},
   "outputs": [],
   "source": [
    "import numpy as np\n",
    "import matplotlib.pyplot as plt"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "9cfe4fd4",
   "metadata": {},
   "outputs": [],
   "source": [
    "def P_l_gen(E_o, R_t):\n",
    "    return lambda R_l: R_l*(E_o/(R_t + R_l))**2"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "45c6459c",
   "metadata": {},
   "source": [
    "#### Problem 1\n",
    "1. What should be the load of a practical source($J_\\mathrm{s}=10\\mathrm{A}$, $G_\\mathrm{t}=1\\mathrm{S}$) , such that the system efficiency is $\\eta \\% = 75\\%$.\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "25c35448",
   "metadata": {},
   "source": [
    "$$ R_\\mathrm{t} = 1\\Omega$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "07cf8ee2",
   "metadata": {},
   "source": [
    "Via equation (3),  \n",
    "\n",
    "$$ \\frac{R_\\mathrm{l}}{R_\\mathrm{t} + R_\\mathrm{l}} \\,100\\%= \\eta \\%$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "248cb1fe",
   "metadata": {},
   "source": [
    "$$ \\frac{R_\\mathrm{l}}{1\\Omega + R_\\mathrm{l}} = 0.75$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3dc3053d",
   "metadata": {},
   "source": [
    "$$ R_\\mathrm{l} = 0.75\\Omega + 0.75R_\\mathrm{l}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "dd0162d9",
   "metadata": {},
   "source": [
    "$$ 0.25R_\\mathrm{l} = 0.75\\Omega$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e5af0aee",
   "metadata": {},
   "source": [
    "$$ R_\\mathrm{l} = 3\\Omega$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a13b4854",
   "metadata": {},
   "source": [
    "#### Problem 2\n",
    "2. A practical linear source has been loaded, first by an ideal voltmeter, then, by ammeter of $R_\\mathrm{A} = 100\\Omega$. The indications are $15 \\mathrm{V}$ and $0.1 \\mathrm{A}$, respectively. Draw $I-U$ relationship -- or, in other words, plot $I(U)$ --\n",
    "    find load resistance and the power transferred at the maximum power transfer condition."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "9d1d5273",
   "metadata": {},
   "source": [
    "$$ E_\\mathrm{o} = 15\\mathrm{V}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c725855d",
   "metadata": {},
   "source": [
    "Via KVL and Ohm's Law,  \n",
    "\n",
    "\\begin{align*}\n",
    "    0 &= U - R_\\mathrm{A}I \\\\\n",
    "     &= 15\\mathrm{V} - R_\\mathrm{t}0.1\\mathrm{A} - 10\\mathrm{V} \\\\\n",
    "     &= 5\\mathrm{V} - R_\\mathrm{t}100\\mathrm{mA}\n",
    "\\end{align*}     "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "56cd247f",
   "metadata": {},
   "source": [
    "$$ R_\\mathrm{t} = 50\\Omega$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ee476d22",
   "metadata": {},
   "source": [
    "\\begin{equation}\\tag{2.3}\n",
    "    U = 15\\mathrm{V} - 50\\Omega\\,I\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f54ba41d",
   "metadata": {},
   "source": [
    "$$ R_\\mathrm{l} = R_\\mathrm{t} = 50\\Omega$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a2dd741c",
   "metadata": {},
   "source": [
    "$$P_\\mathrm{l}^\\max = \\frac{E_\\mathrm{o}^2}{4R_\\mathrm{t}^2} = 22.5\\mathrm{mW}$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "fdaa2f95",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.0225"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "(15**2)/(4*50**2)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a80a0656",
   "metadata": {},
   "source": [
    "$$ I = 300\\mathrm{mA} - 20\\mathrm{mS}\\,U$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "id": "b38041a2",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "15.0"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "300/20"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "id": "a427ea19",
   "metadata": {
    "code_folding": [
     0
    ]
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[<matplotlib.lines.Line2D at 0x7f6aaa43deb0>]"
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     "execution_count": 6,
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    {
     "data": {
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\n",
      "text/plain": [
       "<Figure size 432x288 with 1 Axes>"
      ]
     },
     "metadata": {
      "needs_background": "light"
     },
     "output_type": "display_data"
    }
   ],
   "source": [
    "# Plot I vs U.\n",
    "U = np.linspace(0, 20, num=100)\n",
    "I = 0.3 - 0.02*U\n",
    "\n",
    "fig, ax = plt.subplots()\n",
    "ax.set_title(r'$I$ vs. $U$', size=16)\n",
    "ax.set_xlabel('input voltage ($U$) (volts)')\n",
    "ax.set_ylabel('current (mA)')\n",
    "\n",
    "ax.plot(U, I*1000)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "4096a6fc",
   "metadata": {},
   "source": [
    "#### Problem 3\n",
    "3. A practical source of($E_\\mathrm{o}=10 \\mathrm{V}$, $R_\\mathrm{t}=10\\Omega$), has been loaded by a resistor of variable resistance $R=R_\\mathrm{l}\\in\\mathbb{R}\\; : \\; 2\\le R \\le 8$.  Find the minimum and the maximum power supplied."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "id": "7c46fb87",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(1.388888888888889, 2.469135802469136)"
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "P_l = P_l_gen(10, 10)\n",
    "(P_l(2), P_l(8))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "id": "71c68132",
   "metadata": {
    "code_folding": [
     0
    ]
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[<matplotlib.lines.Line2D at 0x7f6aa82ae550>]"
      ]
     },
     "execution_count": 8,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
      "image/png": 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\n",
      "text/plain": [
       "<Figure size 432x288 with 1 Axes>"
      ]
     },
     "metadata": {
      "needs_background": "light"
     },
     "output_type": "display_data"
    }
   ],
   "source": [
    "# Plot the power supplied.\n",
    "R = R_l = np.linspace(2, 16, num=100)\n",
    "\n",
    "fig, ax = plt.subplots()\n",
    "ax.set_title('Power Supplied', size=16)\n",
    "ax.set_xlabel('Load resistance (ohms)')\n",
    "ax.set_ylabel('Power supplied (watts)')\n",
    "\n",
    "ax.plot(R, P_l(R))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "8c4e4177",
   "metadata": {},
   "source": [
    "#### Problem 4\n",
    "4. A practical source ($E_\\mathrm{o}=4\\mathrm{V}$, $R_\\mathrm{t}=2\\Omega$), has been loaded by a resistor. Find range of its resistance $R:=R_\\mathrm{l}\\in\\mathbb{R}\\; : \\; R_\\min \\le R \\le R_\\max$ so that the power transferred is $0.5\\mathrm{W}\\le P_\\mathrm{l} \\le 1\\mathrm{W}$."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "88a8d0ce",
   "metadata": {},
   "source": [
    "The max power transferred ($P_\\mathrm{l}^\\max$) occurs when $R_\\mathrm{l}=R_\\mathrm{t}=2\\Omega$ and is given by:  \n",
    "\n",
    "$$ P_\\mathrm{l}^\\max = \\frac{E_\\mathrm{o}^2}{4R_\\mathrm{t}^2} = \\frac{4}{4}\\,\\mathrm{W} = 1\\mathrm{W}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "5cf0427e",
   "metadata": {},
   "source": [
    "Which means that, $R_\\max = R_\\mathrm{t} = 2\\Omega$."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "29dcee76",
   "metadata": {},
   "source": [
    "Via equation (1),  \n",
    "\n",
    "\\begin{align*}\n",
    "    P_\\mathrm{l} = 0.5\\mathrm{W} &= \\left( \\frac{E_\\mathrm{o}}{R_\\mathrm{t} + R_\\mathrm{l}}\\right)^2R_\\mathrm{l} \\\\\n",
    "     &= \\left(\\frac{4\\mathrm{V}}{2\\Omega + R_\\min} \\right)^2R_\\min\n",
    "\\end{align*}     "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1635ed92",
   "metadata": {},
   "source": [
    "Multiplying throughout the above equation by $(2\\Omega + R_\\min)^2$ gives:  \n",
    "\n",
    "$$ 0.5\\mathrm{W}\\,(2\\Omega + R_\\min)^2 = 16\\mathrm{V}^2\\,R_\\min$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1fba30d1",
   "metadata": {},
   "source": [
    "$$ 0.5\\mathrm{W}\\,(R_\\min^2 + 4\\Omega \\, R_\\min + 4\\Omega^2) = 16\\mathrm{V}^2\\,R_\\min$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ade9df98",
   "metadata": {},
   "source": [
    "$$ 0.5\\mathrm{W}\\,(R_\\min^2 + 4\\Omega^2) = 14\\mathrm{V}^2\\,R_\\min$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6515197f",
   "metadata": {},
   "source": [
    "Which leaves us with the following quadratic equation,  \n",
    "\n",
    "$$ 0.5\\mathrm{W}\\,R_\\min^2 - 14\\mathrm{V}^2\\,R_\\min + 2\\Omega^2\\mathrm{W} = 0$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1931c273",
   "metadata": {},
   "source": [
    "Dividing throughout by $\\mathrm{W}$ gives:  \n",
    "\n",
    "$$ 0.5R_\\min^2 - 14\\Omega\\,R_\\min + 2\\Omega^2 = 0$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6d33a4c6",
   "metadata": {},
   "source": [
    "Via the quadratic formula,  \n",
    "\n",
    "$$ R_\\min = \\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "id": "3b0acd2d",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(27.856406460551018, 0.14359353944898245)"
      ]
     },
     "execution_count": 9,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a, b, c = 0.5, -14, 2\n",
    "((-b+np.sqrt(b**2 - 4*a*c))/(2*a), (-b-np.sqrt(b**2 - 4*a*c))/(2*a))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f94cab7f",
   "metadata": {},
   "source": [
    "#### Problem 5\n",
    "5. A battery has open-circuit voltage of $9 \\mathrm{V}$ and short-circuit current of $3 \\mathrm{A}$. Find the load\n",
    "    resistance at 25% efficiency of the system."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "9d7ffaa2",
   "metadata": {},
   "source": [
    "$$ E_\\mathrm{o}=9\\mathrm{V} \\quad J_\\mathrm{s} = 3\\mathrm{A} \\quad R_\\mathrm{t} = \\frac{9\\mathrm{V}}{3\\mathrm{A}} = 3\\Omega$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "723161ec",
   "metadata": {},
   "source": [
    "Via equation (1),  \n",
    "\n",
    "$$ P_\\mathrm{l} = \\left( \\frac{E_\\mathrm{o}}{R_\\mathrm{t}+R_\\mathrm{l}}\\right)^2R_\\mathrm{l} = \\left(\\frac{9\\mathrm{V}}{3\\Omega + R_\\mathrm{l}} \\right)^2R_\\mathrm{l}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7986fc48",
   "metadata": {},
   "source": [
    "Via equation (3),  \n",
    "\n",
    "$$ \\eta \\% = 25\\% = \\frac{R_\\mathrm{l}}{3\\Omega + R_\\mathrm{l}}\\,100\\%$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a0fccf93",
   "metadata": {},
   "source": [
    "$$ R_\\mathrm{l} = 0.25(3\\Omega+R_\\mathrm{l})$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "76d8151c",
   "metadata": {},
   "source": [
    "$$ 0.75R_\\mathrm{l} = 0.75\\Omega$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "9212b3b9",
   "metadata": {},
   "source": [
    "$$ R_\\mathrm{l}=1\\Omega$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "cf5f66dc",
   "metadata": {},
   "source": [
    "#### Problem 6\n",
    "6. Find the minimum and the maximum power transferred from a practical source of ($J_\\mathrm{s}=1.5\\mathrm{A}$, $0.25\\mathrm{S}$) to a variable load ranging from $4\\Omega$ to $6\\Omega$."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "2ebef408",
   "metadata": {},
   "source": [
    "$$ R_\\mathrm{t} = 4\\Omega\\quad\\quad\\quad E_\\mathrm{o}=J_\\mathrm{s}R_\\mathrm{t} = 6\\mathrm{V}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ccfc4a7b",
   "metadata": {},
   "source": [
    "$$ R:=R_\\mathrm{l} \\in\\mathbb{R}\\; : \\; 4\\Omega\\le R \\le 6\\Omega$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f0101f4c",
   "metadata": {},
   "source": [
    "The max power transfer occurs when $R_\\mathrm{t}=R_\\mathrm{l} = 4\\Omega$ which is within the range that $R_\\mathrm{l}$ varies within.  Thus,  \n",
    "\n",
    "$$ P_\\mathrm{l}^\\max  = \\frac{9\\mathrm{V}^2}{4\\Omega} = 2.25\\mathrm{W}$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "id": "d6e82c66",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "2.25"
      ]
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "9/4"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "2c57f632",
   "metadata": {},
   "source": [
    "Since the curve given by $P_\\mathrm{l}(R)$ only has one critical point, which occurs at $(R_\\mathrm{t}=R, P_\\mathrm{l}^\\max)$, the minimum value of $P_\\mathrm{l}$, denoted as '$P_\\mathrm{l}^\\min$', within the given domain of $R\\in\\mathbb{R}\\; :\\; 4\\Omega\\le R \\le 6\\Omega$, must therefore occur at $\\big(6\\Omega, P_\\mathrm{l}(6\\Omega)\\big)$."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6699d377",
   "metadata": {},
   "source": [
    "$$ P_\\mathrm{l}^\\min = \\left(\\frac{6\\mathrm{V}}{10\\Omega}\\right)^2 6\\Omega = 2.16\\mathrm{W}$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "id": "a279b658",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "2.16"
      ]
     },
     "execution_count": 11,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "(6/10)**2*6"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "208749ad",
   "metadata": {},
   "source": [
    "#### Problem 7\n",
    "7. Calculate the efficiency of the system presented in Fig. 2.5.14."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "0b0cfd8b",
   "metadata": {},
   "source": [
    "$$ J_\\mathrm{s}=1\\mathrm{mA}\\quad\\quad E_\\mathrm{o}=9\\mathrm{V}\\quad\\quad R_\\mathrm{t} = \\frac{E_\\mathrm{o}}{J_\\mathrm{s}} = 9\\mathrm{k}\\Omega$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "64c25868",
   "metadata": {},
   "source": [
    "Via equation (3),  \n",
    "\n",
    "$$ \\eta \\% = \\frac{R_\\mathrm{l}}{R_\\mathrm{t} + R_\\mathrm{l}}\\,100\\% = \\frac{R_\\mathrm{l}}{9\\mathrm{k}\\Omega + R_\\mathrm{l}}\\,100\\%$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "id": "65fa337e",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[<matplotlib.lines.Line2D at 0x7f6aa81e8a00>]"
      ]
     },
     "execution_count": 12,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
      "image/png": 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\n",
      "text/plain": [
       "<Figure size 432x288 with 1 Axes>"
      ]
     },
     "metadata": {
      "needs_background": "light"
     },
     "output_type": "display_data"
    }
   ],
   "source": [
    "R = R_l = np.linspace(0, 1e5, num=200)\n",
    "eta = 100*R/(9000+R)\n",
    "\n",
    "fig, ax = plt.subplots()\n",
    "ax.set_title(r'Power System Efficiency ($\\eta \\%$)',size=16)\n",
    "ax.set_xlabel(r'Load Resistance ($\\mathrm{k}\\Omega$)')\n",
    "ax.set_ylabel(r'$\\eta\\%$')\n",
    "\n",
    "ax.plot(R/1000, eta)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "133af8d8",
   "metadata": {},
   "source": [
    "#### Problem 8\n",
    "8. Find the maximum power absorbed by the load resistor.  \n",
    "\n",
    "![Fig-P.2.6.8](Figures/Fig-P.2.6.8.png)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "994fbd68",
   "metadata": {},
   "source": [
    "First we convert the left branch consisting of a Thevenin equivalent ($E_1 = 11\\mathrm{V}$, $R_1 = 6\\Omega + 3\\Omega = 9\\Omega$) into it's Norton equivalent."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c9d8848b",
   "metadata": {},
   "source": [
    "$$J_1 = \\frac{11\\mathrm{V}}{9\\Omega} \\approxeq 1.22\\mathrm{A} $$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "id": "253e4c3c",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "1.2222222222222223"
      ]
     },
     "execution_count": 13,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "J_1 = 11/9\n",
    "J_1"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7f5cc2ed",
   "metadata": {},
   "source": [
    "And now we can combine this newly obtained subcircuit with the current source that is in parallel it, denoted as '$J_2$'.  The subcircuit that is obtained by doing so has a short-circuit current ($J_\\mathrm{s}$) of:  \n",
    "\n",
    "$$ J_\\mathrm{s} = J_1 + J_2 \\approxeq 2.22\\mathrm{A}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b5dd0f77",
   "metadata": {},
   "source": [
    "And a conductance of:  \n",
    "\n",
    "$$ G_\\mathrm{t} = \\frac{1}{R_1} = \\frac{1}{9}\\,\\mathrm{S}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "094ac287",
   "metadata": {},
   "source": [
    "The open-circuit voltage of this subcircuit ($E_\\mathrm{o}$) is given by:  \n",
    "\n",
    "$$ E_\\mathrm{o} = J_\\mathrm{s}R_\\mathrm{t} = 20\\mathrm{V}$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "id": "b6700a70",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "20.0"
      ]
     },
     "execution_count": 14,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "E_o = (11/9 + 1)*9\n",
    "E_o"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e163c611",
   "metadata": {},
   "source": [
    "And, we can combine this Thevenin equivalent with the $11\\Omega$ resistance that is in series with it to obtain a subcircuit consisting of ($E_\\mathrm{o}=20\\mathrm{V}$, $R_\\mathrm{t} = 9\\Omega + 11\\Omega = 20\\Omega$)."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ccfb2904",
   "metadata": {},
   "source": [
    "The max power condition is $R_\\mathrm{t}=R_\\mathrm{l}=20\\Omega$ and the max power is given by:  \n",
    "\n",
    "$$ P_\\mathrm{l}^\\max = \\frac{1}{4}\\,\\mathrm{W} = 250\\mathrm{mW}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "2c033340",
   "metadata": {},
   "source": [
    "#### Problem 9\n",
    "9. If a practical current source: $J_\\mathrm{s}=2\\mathrm{A}$, $G_\\mathrm{t}0.5\\mathrm{S}$ and a voltage source: $E_\\mathrm{o}=\\text{?}$, $R_\\mathrm{t}=4\\Omega$ are connected back to back (“+” with “+”), is there a value for $E_\\mathrm{o}$ for which there is no power transfer between them ? If so, find the voltage."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6355f2a7",
   "metadata": {},
   "source": [
    "Let:\n",
    "\\begin{matrix}\n",
    "    J_1:=J_\\mathrm{s}=2\\mathrm{A} \\quad  \\quad &G_1:=G_\\mathrm{t}=0.5\\mathrm{S} &\\\\\n",
    "   E_2:=E_\\mathrm{o}=\\text{?} \\quad  \\quad &R_2:=R_\\mathrm{t}=4\\Omega &\n",
    "\\end{matrix}"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "9848ed36",
   "metadata": {},
   "source": [
    "The answer is yes at it is when $E_1=E_2$."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "5855f372",
   "metadata": {},
   "source": [
    "$$ E_2 = E_1 = J_1R_1 = 4\\mathrm{V}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "fef72968",
   "metadata": {},
   "source": [
    "*Ignore the rest of this problem's solution;  It doesn't provide an answer.*"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a76d11e4",
   "metadata": {},
   "source": [
    "The power delivered to the voltage source from the current source is:  \n",
    "\n",
    "\\begin{align*}\\tag{9.1}\n",
    "    P_2 = IU &= U(J_2 - 0.25\\mathrm{S}\\,U) \\\\\n",
    "     &= 0.25\\mathrm{S}\\cdot U(E_2 - U)\n",
    "\\end{align*}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a7d17034",
   "metadata": {},
   "source": [
    "Assuming $P_2=0\\mathrm{W}$,  \n",
    "\n",
    "$$ E_2 = U$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "9556a90a",
   "metadata": {},
   "source": [
    "Since $U$ is a function of $I$ -- $U:=U(I) = E_2 + R_2I$ -- and $E_2$ is not a function of $I$, $I$, $U$, and $E_2$ would have to equal zero for $P_2$ to be equal to zero, but that is inconsistent with the following,  \n",
    "\n",
    "$$ E_1 = J_1R_1 = 2\\mathrm{A}\\,2\\Omega = 4\\mathrm{V}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "55a10fcb",
   "metadata": {},
   "source": [
    "And, via KVL, and Ohm's law,  \n",
    "\n",
    "$$ E_1 - R_1I - R_2I - E_2 = 4\\mathrm{V} - 6\\Omega\\,I = 0$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "010ff619",
   "metadata": {},
   "source": [
    "$$ I = \\frac{4\\mathrm{V}}{6\\Omega} = \\frac{2}{3}\\,\\mathrm{A}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6887d3dc",
   "metadata": {},
   "source": [
    "Now, what about the other way around?  The power delivered to the current source from the voltage source is given by:  \n",
    "\n",
    "\\begin{align*}\\tag{9.2}\n",
    "    P_1 = IU &= U(-J_1 + 0.5\\mathrm{S}\\,U) \\\\\n",
    "     &= U(0.5\\mathrm{S}\\,U - 2\\mathrm{A})\n",
    "\\end{align*}     "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "100563b3",
   "metadata": {},
   "source": [
    "Assuming $P_1 = 0\\mathrm{W}$,  \n",
    "\n",
    "$$ 0.5\\mathrm{S}\\,U = 2\\mathrm{A}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "4ed8f13f",
   "metadata": {},
   "source": [
    "$$ U = 4\\mathrm{V} = E_2 - 4\\Omega\\, I$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f3c19b57",
   "metadata": {},
   "source": [
    "$$ I = J_2 - G_2U = G_2(E_2 - U) = 0.25\\mathrm{S}\\,(E_2 - 4\\mathrm{V})$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "2db1c060",
   "metadata": {},
   "source": [
    "Via the above two equations and sustitution,  \n",
    "\n",
    "$$ $$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "19b57063",
   "metadata": {},
   "source": [
    "#### Problem 10\n",
    "10. Find $R$ such that maximum power is dissipated in $R$. Calculate its value.  \n",
    "\n",
    "![Fig-P.2.6.10](Figures/Fig-P.2.6.10.png)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f504e00b",
   "metadata": {},
   "source": [
    "The power dissipated in $R$ is given by:  \n",
    "\n",
    "\\begin{equation}\\tag{10.1}\n",
    "    P_R = I_R^2 R\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "69e13739",
   "metadata": {},
   "source": [
    "Via KCL,  \n",
    "\n",
    "$$ 5\\mathrm{A} + I_E - I_R = 0\\mathrm{A}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "46427c25",
   "metadata": {},
   "source": [
    "\\begin{equation}\\tag{10.2}\n",
    "    I_R = 5\\mathrm{A} + I_E\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a6e16ec3",
   "metadata": {},
   "source": [
    "Via KCL and Ohm's Law,  \n",
    "\n",
    "$$U_J - 12\\Omega\\,5\\mathrm{A} - I_R R = U_J - 60\\mathrm{V} - I_R R = 0 $$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ba9085f1",
   "metadata": {},
   "source": [
    "\\begin{equation}\\tag{10.3}\n",
    "    I_R R = U_J - 60\\mathrm{V}\n",
    "\\end{equation}"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "06f1dda7",
   "metadata": {},
   "source": [
    "and,\n",
    "$$ I_R R + 5\\Omega\\, I_E - 10\\mathrm{V} + 5\\Omega\\, I_E = I_R R + 10\\Omega\\, I_E - 10\\mathrm{V} = 0$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "d1b25cc8",
   "metadata": {},
   "source": [
    "\\begin{equation}\\tag{10.4}\n",
    "    10\\mathrm{V} = I_R R + 10\\Omega\\, I_E\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "264e449e",
   "metadata": {},
   "source": [
    "Via equations (10.4) and (10.2) and substitution,  \n",
    "\n",
    "$$ 10\\mathrm{V} = I_R R + 10\\Omega(I_R - 5\\mathrm{A})$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "60e94f26",
   "metadata": {},
   "source": [
    "$$ 60\\mathrm{V} = I_R(R + 10\\Omega)$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "872835e0",
   "metadata": {},
   "source": [
    "\\begin{equation}\\tag{10.5}\n",
    "    I_R = \\frac{60\\mathrm{V}}{R + 10\\Omega}\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "0dff5c27",
   "metadata": {},
   "source": [
    "Via equations (10.5) and (10.1) and substitution, we can obtain the power dissipated in $R$ as a function of $R$.  \n",
    "\n",
    "\\begin{equation}\\tag{10.1a}\n",
    "    P_R := P_R(R) = \\left( \\frac{60\\mathrm{V}}{R + 10\\Omega}\\right)^2R\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "d396e5a2",
   "metadata": {},
   "source": [
    "The above identity fits the form of equation (1), whereby $E_\\mathrm{o} = 60\\mathrm{V}$, $R_\\mathrm{t}=10\\Omega$ and $R_\\mathrm{l}=R$."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "4952d779",
   "metadata": {},
   "source": [
    "Thus, via relationship (2), the maximum power condition is $R = 10\\Omega$ and the maximum power dissipated in $R$ is,  \n",
    "\n",
    "$$ P_R^\\max = \\frac{60^2\\mathrm{V}^2}{4\\cdot10 \\Omega} = 90\\mathrm{W}$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "id": "1ec5134c",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "90.0"
      ]
     },
     "execution_count": 15,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "60**2/40"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "dc132db0",
   "metadata": {},
   "source": [
    "#### Problem 11\n",
    "11. In a circuit of Problem 2.6.10, resistance $R$ varies from $4\\Omega$ to $8\\Omega$. Find the minimum and\n",
    "    the maximum power dissipated by this resistance."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7a89c53a",
   "metadata": {},
   "source": [
    "The minumin power dissipation by $R$ occurs when $R=4\\Omega$.  Said power dissipation is given by:  \n",
    "\n",
    "$$ P_R^\\min = \\left(\\frac{60\\mathrm{V}}{14\\Omega}\\right)^2 4\\Omega \\approxeq 73.47\\mathrm{W}$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "id": "5d8712e2",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "73.46938775510203"
      ]
     },
     "execution_count": 16,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "4*(60/14)**2"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e088b963",
   "metadata": {},
   "source": [
    "and, the maximum power dissipated (within the given load resistance domain) is given by,  \n",
    "\n",
    "$$ P_R^\\max = \\left(\\frac{60}{18} \\right)^2 8\\Omega \\approxeq 88.89\\mathrm{W}$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "id": "1ec0b0e6",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "88.8888888888889"
      ]
     },
     "execution_count": 17,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "8*(60/18)**2"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "10ea5fb2",
   "metadata": {},
   "source": [
    "#### Problem 12\n",
    "12. Find an expression for the maximum power available from two identical sources, each\n",
    "    characterized by emf $E_\\mathrm{o}$ and internal resistance $R_\\mathrm{t}$, if they are connected: a) in series, b) in parallel."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "df777d63",
   "metadata": {},
   "source": [
    "**a)**  The Thevenin equivalent $(E_\\mathrm{o}^*, R_\\mathrm{t}^*)$ of the two sources in series is given by:  \n",
    "\n",
    "$$ E_\\mathrm{o}^* = 2E_\\mathrm{o}\\quad\\quad R_\\mathrm{t}^* = 2R_\\mathrm{t}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ba1ea7b3",
   "metadata": {},
   "source": [
    "The power available from this active subcircuit, in terms of $U$, or as a function of $U$, is:  \n",
    "\n",
    "\\begin{align*}\n",
    "    P :=P(U) &= IU \\\\\n",
    "     &= U\\left(\\frac{E_\\mathrm{o}^*}{R_\\mathrm{t}^*} - \\frac{U}{R_\\mathrm{t}^*}\\right) \\\\ \\\\\n",
    "     P(U)&= \\frac{UE_\\mathrm{o}^* - U^2}{R_\\mathrm{t}^*}\n",
    "\\end{align*}     "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "22aff2bd",
   "metadata": {},
   "source": [
    "We need to find a value $U$ such that,  \n",
    "\n",
    "$$ P'(U)= 0$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "88d5e6ab",
   "metadata": {},
   "source": [
    "$$ E_\\mathrm{o}^* - 2U = 0$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "dacf921b",
   "metadata": {},
   "source": [
    "$$ U = \\frac{E_\\mathrm{o}^*}{2} = E_\\mathrm{o}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "8e200666",
   "metadata": {},
   "source": [
    "**b)** The Norton equivalent $(J_\\mathrm{s}^*, G_\\mathrm{t}^*)$ of the two sources in parallel is given by:  \n",
    "\n",
    "$$ J_\\mathrm{s}^* = 2J_\\mathrm{s} = 2E_\\mathrm{o}G_\\mathrm{t}\\quad \\quad G_\\mathrm{t}^*  = 2G_\\mathrm{t}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "36725f4b",
   "metadata": {},
   "source": [
    "$$ J_\\mathrm{s} = \\frac{E_\\mathrm{o}}{R_\\mathrm{t}} = E_\\mathrm{o}G_\\mathrm{t}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "fc47e230",
   "metadata": {},
   "source": [
    "The power available from this active subcircuit, in terms of $I$, or as a function of $I$, is:  \n",
    "\n",
    "\\begin{align*}\n",
    "    P:=P(I) &= IU \\\\\n",
    "     &= I(R_\\mathrm{t}^* J_\\mathrm{s}^* - R_\\mathrm{t}^* I)\n",
    "\\end{align*}     "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ea9bb40e",
   "metadata": {},
   "source": [
    "$$ P'(I) = R_\\mathrm{t}^* J_\\mathrm{s}^* - 2R_\\mathrm{t}^* I=0$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "53f02547",
   "metadata": {},
   "source": [
    "$$ J_\\mathrm{s}^* - 2I = 0$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "0275eb31",
   "metadata": {},
   "source": [
    "$$I = J_\\mathrm{s}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "8e8d8da9",
   "metadata": {},
   "source": [
    "#### Problem 13\n",
    "13. Two active subcircuits characterized by Thevenin equivalents are as shown. Suppose $E_{\\mathrm{o}1}=12\\mathrm{V}$, $R_{\\mathrm{t}1}=3\\Omega$, $E_{\\mathrm{o}2}=2\\mathrm{V}$ and $R_{\\mathrm{t}2}$ is adjustable. Find the value of $R_{\\mathrm{t}2}$ such that the maximum power is transferred from subcircuit 1 to subcircuit 2. What is the value of this\n",
    "    power ?"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "cfa1d417",
   "metadata": {},
   "source": [
    "Let  \n",
    "\n",
    "$$ E_\\mathrm{o} := E_{\\mathrm{o}1} - E_{\\mathrm{o}2} = 10\\mathrm{V}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "cf89e337",
   "metadata": {},
   "source": [
    "and,\n",
    "$$R_\\mathrm{t}:=R_{\\mathrm{t}1} = 3\\Omega$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "718cca91",
   "metadata": {},
   "source": [
    "The max power transfer condition is $R_\\mathrm{l}:=R_{\\mathrm{t}2} = R_\\mathrm{t} = 3\\Omega$.  The max power transfered is given by:  \n",
    "\n",
    "$$ P_\\mathrm{l}^\\max = \\frac{E_\\mathrm{o}^2}{4\\mathrm{t}} = \\frac{100\\mathrm{V}^2}{12\\Omega}\\approxeq 8.33\\mathrm{W}$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "id": "f34ee988",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "8.333333333333334"
      ]
     },
     "execution_count": 18,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "100/12"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "5e981516",
   "metadata": {},
   "source": [
    "#### Problem 14\n",
    "14. Two active subcircuits characterized by Thevenin equivalents are as shown in Fig. P.2.6.13. Suppose $E_{\\mathrm{o}2}=2\\mathrm{V}$ ,$R_{\\mathrm{t2}} = 2\\Omega$, $R_{\\mathrm{t}1} = 3\\Omega$  and $E_{\\mathrm{o}1}$ is adjustable. Find the value of $E_{\\mathrm{o}1}$ such that the maximum power is transferred from subcircuit 2 to subcircuit 1. What is the value of this power ?"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "2f9b5028",
   "metadata": {},
   "source": [
    "The power transferred to subcircuit 1 is given by:  \n",
    "\n",
    "$$ -P_1 = IU$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "9478183c",
   "metadata": {},
   "source": [
    "\\begin{align*}\\tag{14.3}\n",
    "    -P_1:=-P_1(I) &= I(E_{\\mathrm{o}1} + R_{\\mathrm{t}1}I) \\\\\n",
    "     &= E_\\mathrm{\\mathrm{o}1}I + 3\\Omega\\,I^2\n",
    "\\end{align*}     "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "d752b4d6",
   "metadata": {},
   "source": [
    "We need to find an $I$ such that:  \n",
    "\n",
    "$$P'(I) = E_{\\mathrm{o}1} + 6\\Omega\\,I = 0 $$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "72f78982",
   "metadata": {},
   "source": [
    "\\begin{equation}\\tag{14.1}\n",
    "    I = -\\frac{E_{\\mathrm{o}1}}{6\\Omega}\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "406aaff7",
   "metadata": {},
   "source": [
    "Via KVL and Ohms Law,  \n",
    "\n",
    "$$ E_{\\mathrm{o}1} + 3\\Omega\\,I + 2\\Omega\\,I - 2\\mathrm{V} = 0 $$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "347e86c8",
   "metadata": {},
   "source": [
    "$$ 5\\Omega I = 2\\mathrm{V} - E_{\\mathrm{o}1}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ed52bb6f",
   "metadata": {},
   "source": [
    "\\begin{equation}\\tag{14.2}\n",
    "    I = 400\\mathrm{mA} - 0.2\\Omega^{-1}\\,E_{\\mathrm{o}1}\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e7c9bc5d",
   "metadata": {},
   "source": [
    "Via equations (14.1) and (14.2),  \n",
    "\n",
    "$$ 400\\mathrm{mA} = E_{\\mathrm{o}1}\\left(\\frac{1}{5\\Omega} - \\frac{1}{6\\Omega} \\right)$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7f2437c4",
   "metadata": {},
   "source": [
    "\\begin{equation}\\tag{14.4}\n",
    "    E_{\\mathrm{o}1} = 0.4\\mathrm{A} \\cdot 30\\Omega = 12\\mathrm{V}\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "id": "fc4b98a0",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "12.0"
      ]
     },
     "execution_count": 19,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "0.4*30"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a61ff314",
   "metadata": {},
   "source": [
    "The max power transferred, via equations (14.1) and (14.3), is:  \n",
    "\n",
    "$$ -P_1^\\max = \\frac{E_{\\mathrm{o}1}^2}{12\\Omega} - \\frac{E_{\\mathrm{o}1}^2}{6\\Omega} = -\\frac{E_{\\mathrm{o}1}^2}{12\\Omega}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "65111bf7",
   "metadata": {},
   "source": [
    "#### Problem 15\n",
    "15. A subcircuit characterized by the Thevenin equivalent: $E_\\mathrm{o}=10\\mathrm{V}$, $R_\\mathrm{t}=2.4\\Omega$ is loaded by two resistors, $R_{\\mathrm{l}1}$ and $R_{\\mathrm{l}2}$, connected in parallel. Suppose $R_{\\mathrm{l}1}=6\\Omega$ and $R_{\\mathrm{l}2}$ is adjustable. Find the value of $R_{\\mathrm{l}2}$ such that the maximum power is transferred. What is the value of this power?"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "763e423e",
   "metadata": {},
   "source": [
    "$$ R_\\mathrm{l} = \\frac{1}{G_\\mathrm{l}}\\quad \\text{and} \\quad G_\\mathrm{l} = G_{\\mathrm{l}1} + G_{\\mathrm{l}2} = \\frac{1}{6\\Omega} + \\frac{1}{R_{\\mathrm{l}2}}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "9ba906dd",
   "metadata": {},
   "source": [
    "The power transferred as a function of $R_\\mathrm{l}$ is:  \n",
    "\n",
    "\\begin{equation}\\tag{15.1}\n",
    "    P_\\mathrm{l} := P_\\mathrm{l}(R_\\mathrm{l}) = \\left(\\frac{10\\mathrm{V}}{2.4\\Omega + R_\\mathrm{l}} \\right)^2 R_\\mathrm{l}\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f6e3ae0e",
   "metadata": {},
   "source": [
    "Note that, via the chain rule,  \n",
    "\n",
    "$$ P_\\mathrm{l}'(R_{\\mathrm{l}2}) = P_\\mathrm{l}'(R_\\mathrm{l})\\cdot R_\\mathrm{l}'(R_{\\mathrm{l}2})$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a77e0144",
   "metadata": {},
   "source": [
    "or, to put it another way,  \n",
    "\n",
    "\\begin{equation}\\tag{15.2}\n",
    "    \\frac{\\mathrm{d}P_\\mathrm{l}}{\\mathrm{d}R_{\\mathrm{l}2}} = \\frac{\\mathrm{d}P_\\mathrm{l}}{\\mathrm{d}R_\\mathrm{l}}\\cdot \\frac{\\mathrm{d}R_\\mathrm{l}}{\\mathrm{d}R_{\\mathrm{l}2}}\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "95cfdc41",
   "metadata": {},
   "source": [
    "$$ P_\\mathrm{l}'(R_\\mathrm{l}) = \\left(\\frac{10\\mathrm{V}}{2.4\\Omega + R_\\mathrm{l}} \\right)^2 - 2R_\\mathrm{l}\\,\\frac{100\\mathrm{V}^2}{(2.4\\Omega + R_\\mathrm{l})^3}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7e77ff97",
   "metadata": {},
   "source": [
    "\\begin{equation}\\tag{15.3}\n",
    "    P_\\mathrm{l}'(R_\\mathrm{l}) = \\frac{100\\mathrm{V}^2(2.4\\Omega + R_\\mathrm{l}) - 200\\mathrm{V}^2 R_\\mathrm{l}}{(2.4\\Omega + R_\\mathrm{l})^3} = \\frac{240\\mathrm{V}^2\\Omega - 100\\mathrm{V}^2 R_\\mathrm{l}}{(2.4\\Omega + R_\\mathrm{l})^3}\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a5a2c913",
   "metadata": {},
   "source": [
    "$$ R_\\mathrm{l}'(R_{\\mathrm{l}2}) = -\\frac{1}{G_\\mathrm{l}^2} \\cdot -\\frac{1}{R_{\\mathrm{l}2}^2}= \\frac{1}{G_\\mathrm{l}^2 R_{\\mathrm{l}2}^2}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6fd20e42",
   "metadata": {},
   "source": [
    "\\begin{align*}\n",
    "    G_\\mathrm{l}^2R_{\\mathrm{l}2}^2 &= (G_{\\mathrm{l}1}^2 + 2G_{\\mathrm{l}1}G_{\\mathrm{l}2} + G_{\\mathrm{l}2}^2)R_{\\mathrm{l}2}^2 \\\\\n",
    "     &= G_{\\mathrm{l}1}^2R_{\\mathrm{l}2}^2 + 2G_{\\mathrm{l}1}R_{\\mathrm{l}2} + 1 \\\\\n",
    "     &= (G_{\\mathrm{l}1}R_{\\mathrm{l}2} + 1)^2\n",
    "\\end{align*}     "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6b88508d",
   "metadata": {},
   "source": [
    "Thus,\n",
    "\\begin{align*}\\tag{15.4}\n",
    "    R_\\mathrm{l}'(R_{\\mathrm{l}2}) &= \\frac{1}{G_\\mathrm{l}^2R_{\\mathrm{l}2}^2} \\\\ \\\\\n",
    "     &= \\frac{1}{(G_{\\mathrm{l}1}R_{\\mathrm{l}2} + 1)^2}\n",
    "\\end{align*}     "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "153e74e0",
   "metadata": {},
   "source": [
    "We need to solve for $R_{\\mathrm{l}2}$ in the following equation:  \n",
    "\n",
    "$$ P_\\mathrm{l}'(R_{\\mathrm{l}2}) = 0$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "24c67a25",
   "metadata": {},
   "source": [
    "Multiplying throughout by the denominator gives:  \n",
    "\n",
    "\\begin{equation}\\tag{15.5}\n",
    "    R_\\mathrm{l} = 2.4\\Omega\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "555b4c0d",
   "metadata": {},
   "source": [
    "Multiplying throughout the above equation by $G_\\mathrm{l}$ gives:  \n",
    "\n",
    "$$ 1 = 2.4\\Omega\\,\\left(\\frac{1}{6\\Omega} + \\frac{1}{R_{\\mathrm{l}2}}\\right)$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3fbcbfb0",
   "metadata": {},
   "source": [
    "$$ \\frac{2.4\\Omega}{R_{\\mathrm{l}2}} = 0.6$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "id": "51465a47",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "4.0"
      ]
     },
     "execution_count": 21,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "2.4/0.6"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "263109ca",
   "metadata": {},
   "source": [
    "\\begin{equation}\\tag{15.6}\n",
    "    R_{\\mathrm{l}2} = 4\\Omega\n",
    "\\end{equation}    "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "193fddf8",
   "metadata": {},
   "source": [
    "$$ P_l^\\max = \\frac{E_\\mathrm{o}^2}{4R_\\mathrm{t}} = \\frac{25\\mathrm{V}^2}{2.4\\Omega} \\approxeq 10.42\\mathrm{W} $$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "id": "8af8acc5",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "10.416666666666668"
      ]
     },
     "execution_count": 23,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "25/2.4"
   ]
  }
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