{ "cells": [ { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "**************\n", "Basic Concepts\n", "**************" ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Charge and Current\n", "##################\n", "\n", ".. math::\n", "\n", " \\newcommand{\\evalAt}[3]{\\left. #1 \\right\\rvert_{#2}^{#3}}\n", " \\newcommand{\\kilo}{\\mathrm{k}}\n", " \\newcommand{\\milli}{\\mathrm{m}}\n", " \\newcommand{\\micro}{\\mu}\n", " \\newcommand{\\nano}{\\mathrm{n}}\n", " \\newcommand{\\pico}{\\mathrm{p}}\n", " \\newcommand{\\ampere}{\\mathrm{A}}\n", " \\newcommand{\\coulomb}{\\mathrm{C}}\n", " \\newcommand{\\second}{\\mathrm{s}}" ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.1\n", "============\n", "\n", "There are :math:`\\frac{1}{1.602 \\times 10^{-19}} = 6.24 \\times 10^{18}` electrons\n", "per :math:`1 \\coulomb` of charge.\n", "\n", "(a)\n", "---\n", "\n", ".. math::\n", "\n", " \\left( 6.482 \\times 10^{17} \\right) \\left( -1.602 \\times 10^{-19} \\right) =\n", " -103.84 \\milli\\coulomb\n", "\n", "(b)\n", "---\n", "\n", ".. math::\n", "\n", " \\left( 1.24 \\times 10^{18} \\right) \\left( -1.602 \\times 10^{-19} \\right) =\n", " -198.65 \\milli\\coulomb\n", "\n", "(c)\n", "---\n", "\n", ".. math::\n", "\n", " \\left( 2.46 \\times 10^{19} \\right) \\left( -1.602 \\times 10^{-19} \\right) =\n", " -3.941 \\coulomb\n", "\n", "(d)\n", "---\n", "\n", ".. math::\n", "\n", " \\left( 1.628 \\times 10^{20} \\right) \\left( -1.602 \\times 10^{-19} \\right) =\n", " -26.08 \\coulomb" ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.2\n", "============\n", "\n", "(a)\n", "---\n", "\n", ".. math::\n", "\n", " i = \\frac{dq}{dt}\n", " = \\frac{d}{dt} \\left( 3t + 8 \\right)\n", " = 3 \\milli\\ampere\n", "\n", "(b)\n", "---\n", "\n", ".. math::\n", "\n", " i = \\frac{dq}{dt}\n", " = \\frac{d}{dt} \\left( 8t^2 + 4t - 2 \\right)\n", " = \\left( 16t + 4 \\right) \\ampere\n", "\n", "(c)\n", "---\n", "\n", ".. math::\n", "\n", " i = \\frac{dq}{dt}\n", " = \\frac{d}{dt} \\left( 3e^{-t} - 5e^{-2t} \\right)\n", " = \\left( -3e^{-t} + 10e^{-2t} \\right) \\nano\\ampere\n", "\n", "(d)\n", "---\n", "\n", ".. math::\n", "\n", " i = \\frac{dq}{dt}\n", " = \\frac{d}{dt} \\left( 10 \\sin 120 \\pi t \\right)\n", " = \\left( 1200 \\pi \\cos 120 \\pi t \\right) \\pico\\ampere\n", "\n", "(e)\n", "---\n", "\n", ".. math::\n", "\n", " i = \\frac{dq}{dt}\n", " = \\frac{d}{dt} \\left( 20 e^{-4t} \\cos 50t \\right)\n", " &= -80 e^{-4t} \\cos 50t - 1000 e^{-4t} \\sin 50t\\\\\n", " &= -40 e^{-4t} \\left( 2 \\cos 50t +50 \\sin 50t \\right) \\micro\\ampere" ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.3\n", "============\n", "\n", "(a)\n", "---\n", "\n", "Given :math:`i(t) = 3 \\ampere` and :math:`q(0) = 1 \\coulomb`,\n", "\n", ".. math::\n", "\n", " q(t) = \\int i dt\n", " = 3t + C\n", "\n", "where\n", "\n", ".. math::\n", "\n", " q(0) &= 3(0) + C\\\\\n", " 1 &= C.\n", "\n", "(b)\n", "---\n", "\n", "Given :math:`i(t) = (2t + 5) \\milli\\ampere` and :math:`q(0) = 0`,\n", "\n", ".. math::\n", "\n", " q(t) = \\int (2t + 5) dt\n", " = t^2 + 5t + C\n", "\n", "where\n", "\n", ".. math::\n", "\n", " q(0) &= (0)^2 + 5(0) + C\\\\\n", " 0 &= C.\n", "\n", "(c)\n", "---\n", "\n", "Given :math:`i(t) = 20 \\cos\\left( 10t + \\pi / 6 \\right) \\micro\\ampere` and\n", ":math:`q(0) = 2 \\micro\\coulomb`,\n", "\n", ".. math::\n", "\n", " q(t) = \\int i dt\n", " = 2 \\sin\\left( 10t + \\pi / 6 \\right) + C\n", "\n", "where\n", "\n", ".. math::\n", "\n", " q(0) &= 2 \\sin\\left( 10(0) + \\pi / 6 \\right) + C\\\\\n", " 2 &= 2 \\sin \\frac{\\pi}{6} + C\\\\\n", " 1 &= C.\n", "\n", "(d)\n", "---\n", "\n", "Given :math:`i(t) = 10e^{-30t} \\sin\\left( 40t \\right) \\ampere` and\n", ":math:`q(0) = 0`,\n", "\n", ".. math::\n", "\n", " q(t) = \\int i dt\n", " = -\\frac{1}{25} e^{-30t} \\left( 3 \\sin 40t + 4 \\cos 40t \\right) + C\n", "\n", "where\n", "\n", ".. math::\n", "\n", " q(0) &= -\\frac{1}{25} e^{-30(0)} \\left(\n", " 3 \\sin 40 (0) + 4 \\cos 40 (0)\n", " \\right) + C\\\\\n", " 0 &= -\\frac{1}{25} 4 + C\\\\\n", " \\frac{4}{25} &= C." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.4\n", "============\n", "\n", "Suppose a current of :math:`7.4 \\ampere` flows through a conductor. The\n", "amount of charge passing through any cross-section of the conductor in\n", ":math:`20 \\second` is\n", "\n", ".. math::\n", "\n", " Q = \\int_{t_0}^t i dt = \\evalAt{it}{0}{20} = 148 \\coulomb." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.5\n", "============\n", "\n", "The total charge transferred over the time interval of\n", ":math:`0 \\leq t \\leq 10 \\second` when\n", ":math:`i(t) = \\frac{1}{2} t \\ampere` is\n", "\n", ".. math::\n", "\n", " Q = \\int_{t_0}^t i dt\n", " = \\left. \\frac{t^2}{4} \\right\\rvert_0^{10}\n", " = 25 \\coulomb." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.6\n", "============\n", "\n", "The charge (:math:`\\milli\\coulomb`) entering a certain element is defined as\n", "\n", ".. math::\n", "\n", " q(t) = \\begin{cases}\n", " 15t & \\text{if } 0 \\leq t \\leq 2\\\\\n", " 30 & \\text{if } 2 \\leq t \\leq 8\\\\\n", " 30 - 7.5 (t - 8) & \\text{if } 8 \\leq t \\leq 12\\\\\n", " 0 & \\text{otherwise.}\n", " \\end{cases}\n", "\n", "(a)\n", "---\n", "\n", "At :math:`t = 1 \\milli\\second`, the current is\n", "\n", ".. math::\n", "\n", " i(t) = \\frac{dq}{dt} = 15 \\ampere.\n", "\n", "(b)\n", "---\n", "\n", "At :math:`t = 6 \\milli\\second`, the current is\n", "\n", ".. math::\n", "\n", " i(t) = \\frac{dq}{dt} = 0 \\ampere.\n", "\n", "(b)\n", "---\n", "\n", "At :math:`t = 10 \\milli\\second`, the current is\n", "\n", ".. math::\n", "\n", " i(t) = \\frac{dq}{dt} = -7.5 \\ampere." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.7\n", "============\n", "\n", "The charge (:math:`\\coulomb`) flowing in a wire is defined as\n", "\n", ".. math::\n", "\n", " q(t) = \\begin{cases}\n", " 25t & \\text{if } 0 \\leq t \\leq 2\\\\\n", " 50 - 25 (t - 2) & \\text{if } 2 \\leq t \\leq 4\\\\\n", " -25 (t - 4) & \\text{if } 4 \\leq t \\leq 6\\\\\n", " -50 + 25 (t - 6) & \\text{if } 6 \\leq t \\leq 8\\\\\n", " 0 & \\text{otherwise.}\n", " \\end{cases}\n", "\n", "The corresponding current (:math:`\\ampere`) is\n", "\n", ".. math::\n", "\n", " i(t) = \\frac{dq}{dt}\n", " = \\begin{cases}\n", " 25 & \\text{if } 0 \\leq t \\leq 2\\\\\n", " -25 & \\text{if } 2 \\leq t \\leq 6\\\\\n", " 25 & \\text{if } 6 \\leq t \\leq 8\\\\\n", " 0 & \\text{otherwise.}\n", " \\end{cases}" ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.8\n", "============\n", "\n", "The current (:math:`\\milli\\ampere`) flowing past a point in a device is\n", "characterized as\n", "\n", ".. math::\n", "\n", " i(t) = \\begin{cases}\n", " 10t & \\text{if } 0 \\leq t \\leq 1 \\milli\\second\\\\\n", " 10 & \\text{if } 1 \\leq t \\leq 2 \\milli\\second\\\\\n", " 0 & \\text{otherwise.}\n", " \\end{cases}\n", "\n", "The total charge through the point is\n", "\n", ".. math::\n", "\n", " Q(t) = \\int_{t_0 = 0}^{t = 2} i dt\n", " = \\int_0^1 10t dt + \\int_1^2 10 dt\n", " = \\evalAt{5t^2}{0}{1} + \\evalAt{10t}{1}{2}\n", " = 15 \\micro\\coulomb." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.9\n", "============\n", "\n", "The current (:math:`\\ampere`) through an element is characterized by\n", "\n", ".. math::\n", "\n", " i(t) = \\begin{cases}\n", " i_1 = 10 & \\text{if } 0 \\leq t \\leq 1\\\\\n", " i_2 = 10 - 5 (t - 1) & \\text{if } 1 \\leq t \\leq 2\\\\\n", " i_3 = 5 & \\text{if } 2 \\leq t \\leq 4\\\\\n", " i_4 = 5 - 5 (t - 4) & \\text{if } 4 \\leq t \\leq 5\\\\\n", " 0 & \\text{otherwise.}\n", " \\end{cases}\n", "\n", "(a)\n", "---\n", "\n", "The total charge that passed through the element at :math:`t = 1 \\second` is\n", "\n", ".. math::\n", "\n", " Q(t) = \\int_{t_0 = 0}^{t = 1} i dt\n", " = \\int_0^1 i_1 dt\n", " = \\evalAt{10t}{0}{1}\n", " = 10 \\coulomb.\n", "\n", "(b)\n", "---\n", "\n", "The total charge that passed through the element at :math:`t = 3 \\second` is\n", "\n", ".. math::\n", "\n", " Q(t) = \\int_{t_0 = 0}^{t = 3} i dt\n", " = \\int_0^1 i_1 dt + \\int_1^2 i_2 dt + \\int_2^3 i_3 dt\n", " = Q(1) +\n", " \\evalAt{10t - \\frac{5}{2} t^2 + 5t}{1}{2} +\n", " \\evalAt{5t}{2}{3}\n", " = 22.5 \\coulomb.\n", "\n", "(c)\n", "---\n", "\n", "The total charge that passed through the element at :math:`t = 5 \\second` is\n", "\n", ".. math::\n", "\n", " Q(t) = \\int_{t_0 = 0}^{t = 5} i dt\n", " = \\int_0^1 i_1 dt + \\int_1^2 i_2 dt + \\int_2^4 i_3 dt + \\int_4^5 i_4 dt\n", " = Q(3) + \\evalAt{5t}{3}{4} +\n", " \\evalAt{5t - \\frac{5}{2} t^2 + 20t}{4}{5}\n", " = 30 \\coulomb." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Voltage, Power, and Energy\n", "##########################\n", "\n", ".. math::\n", "\n", " \\newcommand{\\joule}{\\mathrm{J}}\n", " \\newcommand{\\volt}{\\mathrm{V}}\n", " \\newcommand{\\watt}{\\mathrm{W}}\n", " \\newcommand{\\hour}{\\mathrm{h}}" ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.10\n", "=============\n", "\n", "A lightning bolt with :math:`10 \\kilo\\ampere` strikes an object for\n", ":math:`15 \\micro\\second`. The amount of charge deposited on the object is\n", "\n", ".. math::\n", "\n", " Q = \\int i dt\n", " = \\left( 10 \\times 10^3 \\ampere \\right)\n", " \\left( 15 \\times 10^{-6} \\second \\right)\n", " = 0.15 \\coulomb." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.11\n", "=============\n", "\n", "A rechargeable flashlight battery can deliver :math:`90 \\milli\\ampere` for about\n", ":math:`12 \\hour`. The amount of charge it can release at that rate is\n", "\n", ".. math::\n", "\n", " Q = \\int i dt\n", " = \\evalAt{\\left( 90 \\times 10^{-3} \\right) t}{0}{12 \\times 3600}\n", " = 3.888 \\kilo\\coulomb.\n", "\n", "If the terminal voltage is :math:`1.5 \\volt`, the battery can delivery\n", "\n", ".. math::\n", "\n", " w &= \\int v dq\n", " & \\quad \\text{(1.3)}\\\\\n", " &= \\int v i dt\n", " & \\quad \\text{(1.1), (1.9)}\\\\\n", " &= \\evalAt{\\left( 1.5 \\times 0.09 \\right) t}{0}{12 \\times 3600}\\\\\n", " &= 5.832 \\kilo\\joule\\\\\n", " &= 1.62 \\watt\\hour." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.12\n", "=============\n", "\n", "The current (:math:`\\ampere`) through an element is given by\n", "\n", ".. math::\n", "\n", " i(t) = \\begin{cases}\n", " 3t & \\text{if } 0 \\leq t < 6\\\\\n", " 18 & \\text{if } 6 \\leq t < 10\\\\\n", " -12 & \\text{if } 10 \\leq t < 15\\\\\n", " 0 & \\text{if } t \\geq 15.\n", " \\end{cases}\n", "\n", "The charge (:math:`\\coulomb`) stored in the element over\n", ":math:`0 < t < 20 \\second` is given by\n", "\n", ".. math::\n", "\n", " Q(t) = \\begin{cases}\n", " \\frac{3}{2} t^2 & \\text{if } 0 \\leq t < 6\\\\\n", " 54 + 18 (t - 6) & \\text{if } 6 \\leq t < 10\\\\\n", " 126 - 12 (t - 10) & \\text{if } 10 \\leq t < 15\\\\\n", " 66 & \\text{if } t \\geq 15.\n", " \\end{cases}" ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.13\n", "=============\n", "\n", "Given the charge (:math:`\\milli\\coulomb`) entering the positive terminal of an\n", "element is\n", "\n", ".. math::\n", "\n", " q = 5 \\sin 4 \\pi t,\n", "\n", "the current (:math:`\\milli\\ampere`) is\n", "\n", ".. math::\n", "\n", " i = \\frac{dq}{dt}\n", " = 20 \\pi \\cos 4 \\pi t.\n", "\n", "Suppose the voltage (:math:`\\volt`) across the element (passive sign convention)\n", "is\n", "\n", ".. math::\n", "\n", " v = 3 \\cos 4 \\pi t.\n", "\n", "(a)\n", "---\n", "\n", "The power delivered to the element at :math:`t = 0.3 \\second` is\n", "\n", ".. math::\n", "\n", " p &= vi & \\quad \\text{(1.7)}\\\\\n", " &= 60 \\pi \\cos^2 4 \\pi t\\\\\n", " &= 30 \\pi \\left( 1 + \\cos 8 \\pi t \\right)\n", " & \\quad \\text{cosine identity} \\cos 2 \\alpha = 2 \\cos^2 \\alpha - 1\\\\\n", " &= 123.37 \\milli\\watt.\n", "\n", "(b)\n", "---\n", "\n", "The energy delivered to the element between :math:`0` and :math:`0.6 \\second` is\n", "\n", ".. math::\n", "\n", " w &= \\int_{t_0}^t p dt\n", " & \\quad \\text{(1.9)}\\\\\n", " &= \\evalAt{30 \\pi t + \\frac{15}{4} \\sin 8 \\pi t}{0}{0.6}\\\\\n", " &= 58.76 \\milli\\joule." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.14\n", "=============\n", "\n", "The voltage (:math:`\\volt`) across a device and the current\n", "(:math:`\\milli\\ampere`) through it are\n", "\n", ".. math::\n", "\n", " v(t) = 10 \\cos 2 t\n", " \\quad \\text{and} \\quad\n", " i(t) = 20 \\left( 1 - \\exp -\\frac{t}{2} \\right).\n", "\n", "(a)\n", "---\n", "\n", "The total charge in the device at :math:`t = 1 \\second` is\n", "\n", ".. math::\n", "\n", " Q &= \\int_{t_0 = 0}^t i dt\n", " & \\quad \\text{(1.2)}\\\\\n", " &= \\evalAt{20t + 40 \\exp -\\frac{t}{2}}{0}{1}\\\\\n", " &= 4.26 \\milli\\coulomb.\n", "\n", "(b)\n", "---\n", "\n", "The power consumed by the device at :math:`t = 1 \\second` is\n", "\n", ".. math::\n", "\n", " p &= vi\n", " & \\quad \\text{(1.7)}\\\\\n", " &= 200 \\left( 1 - \\exp -\\frac{t}{2} \\right) \\cos 2t\\\\\n", " &= -32.75 \\milli\\watt." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.15\n", "=============\n", "\n", "The current (:math:`\\milli\\ampere`) entering the positive terminal of a device\n", "is :math:`i(t) = 6 e^{-2t}` and the voltage (:math:`\\milli\\volt`) across the\n", "device is\n", "\n", ".. math::\n", "\n", " v(t) = 10 \\frac{di}{dt} = -120 \\exp -2t.\n", "\n", "(a)\n", "---\n", "\n", "The total charge delivered to the device between :math:`t = 0` and\n", ":math:`t = 2 \\second` is\n", "\n", ".. math::\n", "\n", " Q &= \\int_{t_0 = 0}^t i dt\n", " & \\quad \\text{(1.2)}\\\\\n", " &= \\evalAt{-3 \\exp -2t}{0}{2}\\\\\n", " &= 2.95 \\milli\\coulomb.\n", "\n", "(b)\n", "---\n", "\n", "The power (:math:`\\micro\\watt`) absorbed is\n", "\n", ".. math::\n", "\n", " p &= vi\n", " & \\quad \\text{(1.7)}\\\\\n", " &= -720 \\exp -4t.\n", "\n", "(c)\n", "---\n", "\n", "The energy absorbed in :math:`3 \\second` is\n", "\n", ".. math::\n", "\n", " w &= \\int_{t_0 = 0}^t p dt\n", " & \\quad \\text{(1.9)}\\\\\n", " &= \\evalAt{180 \\exp -4t}{0}{3}\\\\\n", " &= -180 \\micro\\joule." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Circuit Elements\n", "################" ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.16\n", "=============\n", "\n", "The current (:math:`\\milli\\ampere`) through and the voltage (:math:`\\volt`)\n", "across an element are given by\n", "\n", ".. math::\n", "\n", " i(t) = \\begin{cases}\n", " 30t & \\text{if } 0 \\leq t < 2\\\\\n", " 60 - 30 (t - 2) & \\text{if } 2 \\leq t < 4\\\\\n", " 0 & \\text{otherwise.}\n", " \\end{cases}\n", " \\qquad \\text{and} \\qquad\n", " v(t) = \\begin{cases}\n", " 5 & \\text{if } 0 \\leq t < 2\\\\\n", " -5 & \\text{if } 2 \\leq t < 4\\\\\n", " 0 & \\text{otherwise.}\n", " \\end{cases}\n", "\n", "(a)\n", "---\n", "\n", "The power (:math:`\\milli\\watt`) delivered to the element for :math:`t > 0` is\n", "given by\n", "\n", ".. math::\n", "\n", " p(t) = \\begin{cases}\n", " 150t & \\text{if } 0 \\leq t < 2\\\\\n", " -300 + 150 (t - 2) & \\text{if } 2 \\leq t < 4\\\\\n", " 0 & \\text{otherwise.}\n", " \\end{cases}\n", "\n", "(b)\n", "---\n", "\n", "The total energy absorbed by the element for the period of\n", ":math:`0 < t < 4 \\second` is\n", "\n", ".. math::\n", "\n", " w &= \\int_{t_0 = 0}^t p dt\n", " & \\quad \\text{(1.9)}\\\\\n", " &= \\int_0^2 p dt + \\int_2^4 p dt\\\\\n", " &= \\evalAt{75t^2}{0}{2} + \\evalAt{-600t + 75t^2}{2}{4}\\\\\n", " &= 0 \\milli\\joule." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.17\n", "=============\n", "\n", "The amount of power absorbed by element 3 is\n", "\n", ".. math::\n", "\n", " p_1 + p_2 + p_3 + p_4 + p_5 &= 0 & \\quad \\text{(1.8)}\\\\\n", " p_3 &= 205 - 60 - 45 - 30\\\\\n", " &= 70 \\watt." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.18\n", "=============\n", "\n", "Given :math:`I = 10 \\ampere`,\n", "\n", ".. math::\n", "\n", " p_1 &= -\\left( 30 \\times 10 \\right) = -300 \\watt\\\\\n", " \\\\\n", " p_2 &= I \\times 10 = 100 \\watt\\\\\n", " \\\\\n", " p_3 &= 20 \\times 14 = 280 \\watt\\\\\n", " \\\\\n", " p_4 &= -\\left( 8 \\times 4 \\right) = -32 \\watt\\\\\n", " \\\\\n", " p_5 &= -\\left( 12 \\times 0.4I \\right) = -48 \\watt." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.19\n", "=============\n", "\n", "The power (:math:`\\watt`) absorbed by each element is given by\n", "\n", ".. math::\n", "\n", " p_1 &= - \\left( 9 \\times 8 \\right) = -72\\\\\n", " \\\\\n", " p_2 &= 9 \\times 2 = 18\\\\\n", " \\\\\n", " p_3 &= 3I\\\\\n", " \\\\\n", " p_4 &= 6I\n", "\n", "where\n", "\n", ".. math::\n", "\n", " 0 &= \\sum_i p_i\n", " & \\quad \\text{(1.8)}\\\\\n", " &= -54 + 9I\\\\\n", " I &= 6 \\ampere." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.20\n", "=============\n", "\n", "Given :math:`I_o = 2 \\ampere`, the power (:math:`\\watt`) absorbed by each\n", "element is given by\n", "\n", ".. math::\n", "\n", " p_1 &= - \\left( 30 \\times 6 \\right) = -180\\\\\n", " \\\\\n", " p_2 &= 12 \\times 6 = 72\\\\\n", " \\\\\n", " p_3 &= 3 V_o\\\\\n", " \\\\\n", " p_4 &= 28 \\times I_o = 56\\\\\n", " p_5 &= 28 \\times 1 = 28\\\\\n", " p_6 &= -\\left( 5 I_o \\times 3 \\right) = -30\\\\\n", "\n", "where\n", "\n", ".. math::\n", "\n", " 0 &= \\sum_i p_i\n", " & \\quad \\text{(1.8)}\\\\\n", " &= -54 + 3 V_o\\\\\n", " V_o &= 18 \\volt." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Applications\n", "############\n", "\n", ".. math::\n", "\n", " \\newcommand{\\minute}{\\mathrm{min}}\n", " \\newcommand{\\per}{\\mathrm{/}}\n", " \\newcommand{\\cent}{\\mathrm{cents}}\n", " \\newcommand{\\day}{\\mathrm{day}}\n", " \\newcommand{\\watthour}{\\mathrm{Wh}}\n", " \\newcommand{\\amperehour}{\\mathrm{Ah}}\n", " \\newcommand{\\mega}{\\mathrm{M}}" ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.21\n", "=============\n", "\n", "A :math:`60 \\watt` incandescent bulb operates at :math:`120 \\volt`. The amount\n", "of coulombs flowing through the bulb in one day is\n", "\n", ".. math::\n", "\n", " Q &= \\int \\frac{p}{v} dt\n", " & \\quad \\text{(1.6)}\\\\\n", " &= \\frac{60}{120} \\times 3600\\\\\n", " &= 43.2 \\kilo\\coulomb,\n", "\n", "which is equivalent to saying the number of electrons is\n", "\n", ".. math::\n", "\n", " N_e = Q \\left( 6.24 \\times 10^{18} \\right) = 2.696 \\times 10^{23}." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.22\n", "=============\n", "\n", "A lightning bolt strikes an airplane with :math:`40 \\kilo\\ampere` for\n", ":math:`1.2 \\milli\\second`. The charge deposited on the plane is\n", "\n", ".. math::\n", "\n", " Q &= \\int_{t_0 = 0}^t i dt\n", " & \\quad \\text{(1.2)}\\\\\n", " &= \\evalAt{40000t}{0}{0.0012}\\\\\n", " &= 48 \\coulomb." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.23\n", "=============\n", "\n", "A :math:`1.8 \\kilo\\watt` electric heater takes :math:`15 \\minute` to boil a\n", "quantity of water i.e.\n", "\n", ".. math::\n", "\n", " w &= \\int_{t_0 = 0}^{t} p dt\n", " & \\quad \\text{(1.9)}\\\\\n", " &= \\evalAt{1800t}{0}{15 \\times 60}\\\\\n", " &= 1620000 \\joule\\\\\n", " &= 450 \\watthour\n", " & \\quad 1 \\watthour = 3600 \\joule\\\\\n", " &= 0.45 \\kilo\\watthour.\n", "\n", "Suppose this is done once a day and power costs\n", ":math:`10 \\cent\\per\\kilo\\watthour`. The cost of its\n", "operation for :math:`30` days is\n", "\n", ".. math::\n", "\n", " c = 30 \\times 10 \\times w = 135 \\cent." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.24\n", "=============\n", "\n", "A utility company charges :math:`8.2 \\cent\\per\\kilo\\watthour`. Suppose a\n", "consumer operates a :math:`60 \\watt` light bulb continuously for one day i.e.\n", "\n", ".. math::\n", "\n", " w &= \\int_{t_0 = 0}^{t} p dt\n", " & \\quad \\text{(1.9)}\\\\\n", " &= \\evalAt{60t}{0}{24}\\\\\n", " &= 1.44 \\kilo\\watthour.\n", "\n", "That consumer would pay\n", "\n", ".. math::\n", "\n", " c = 8.2 \\times w = 11.81 \\cent." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.25\n", "=============\n", "\n", "A :math:`1.5 \\kilo\\watt` toaster takes roughly :math:`3.5` minutes to heat four\n", "slices of bread i.e.\n", "\n", ".. math::\n", "\n", " w &= \\int_{t_0 = 0}^{t} p dt\n", " & \\quad \\text{(1.9)}\\\\\n", " &= \\evalAt{1500t}{0}{3.5 \\times 60}\\\\\n", " &= 315000 \\joule\\\\\n", " &= 0.0875 \\kilo\\watthour.\n", "\n", "Assuming energy costs :math:`8.2 \\cent\\per\\kilo\\watthour`, the cost of\n", "operating the toaster once per day for one month is\n", "\n", ".. math::\n", "\n", " c = 8.2 \\times 30 \\times w = 21.53 \\cent." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.26\n", "=============\n", "\n", "A flashlight battery has a rating of :math:`0.8 \\amperehour` and a lifetime of\n", ":math:`10 \\hour`.\n", "\n", "(a)\n", "---\n", "\n", "The battery can deliver\n", "\n", ".. math::\n", "\n", " i = \\frac{0.8}{10} = 80 \\milli\\ampere.\n", "\n", "(b)\n", "---\n", "\n", "If its terminal voltage is :math:`6 \\volt`, the battery can give\n", "\n", ".. math::\n", "\n", " p &= vi\n", " & \\quad \\text{(1.7)}\\\\\n", " &= 6 \\times 0.08\\\\\n", " &= 0.48 \\watt.\n", "\n", "(c)\n", "---\n", "\n", "The amount of energy stored in the battery is\n", "\n", ".. math::\n", "\n", " w &= \\int_{t_0 = 0}^{t = 10} p\n", " & \\quad \\text{(1.9)}\\\\\n", " &= \\evalAt{pt}{0}{10}\\\\\n", " &= 4.8 \\watthour." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.27\n", "=============\n", "\n", "A typical automotive battery requires a constant current of :math:`3 \\ampere`\n", "for 4 hours to become fully charged. Suppose the terminal voltage\n", "(:math:`\\volt`) is :math:`10 + \\frac{t}{2}` where :math:`t` is in hours.\n", "\n", "(a)\n", "---\n", "\n", "The amount of charge transported is\n", "\n", ".. math::\n", "\n", " Q &= \\int_{t_0}^t i dt\n", " & \\quad \\text{(1.2)}\\\\\n", " &= \\evalAt{3t}{0}{4 \\times 3600}\\\\\n", " &= 43.2 \\kilo\\coulomb.\n", "\n", "(b)\n", "---\n", "\n", "The terminal voltage reformulated in seconds is :math:`10 + \\frac{0.5t}{3600}`.\n", "The energy expended is\n", "\n", ".. math::\n", "\n", " w &= \\int_{t_0}^t vi dt\n", " & \\quad \\text{(1.9)}\\\\\n", " &= i \\left(\n", " \\evalAt{10t + \\frac{0.25 t^2}{3600}}{0}{4 \\times 3600}\n", " \\right)\\\\\n", " &= 475.2 \\kilo\\joule\\\\\n", " &= 0.132 \\kilo\\watthour\n", " & \\quad 1 \\kilo\\watthour = 3600 \\kilo\\joule.\n", "\n", "(c)\n", "---\n", "\n", "Assuming electricity costs :math:`9 \\cent\\per\\kilo\\watthour`, the charging cost\n", "is\n", "\n", ".. math::\n", "\n", " c = 9 \\times w = 1.188 \\cent." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.28\n", "=============\n", "\n", "A :math:`60 \\watt` incandescent lamp is connected to a :math:`120 \\volt` source.\n", "\n", "(a)\n", "---\n", "\n", "The current through the lamp is\n", "\n", ".. math::\n", "\n", " i &= \\frac{p}{v}\n", " & \\quad \\text{(1.7)}\\\\\n", " &= 0.5 \\ampere.\n", "\n", "(b)\n", "---\n", "\n", "The energy consumed in a non-leap year is\n", "\n", ".. math::\n", "\n", " w &= \\int_{t_0}^t p dt\n", " & \\quad \\text{(1.9)}\\\\\n", " &= \\evalAt{pt}{0}{365 \\times 24}\\\\\n", " &= 525.6 \\kilo\\joule.\n", "\n", "If electricity costs :math:`9 \\cent\\per\\kilo\\watthour`, the cost of operating\n", "the light is\n", "\n", ".. math::\n", "\n", " c = 9 \\times w = \\$47.304." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.29\n", "=============\n", "\n", "An electric stove with four burners and an oven is used as follows:\n", "\n", "- Burner 1: 20 minutes.\n", "- Burner 2: 40 minutes.\n", "- Burner 3: 15 minutes.\n", "- Burner 4: 45 minutes.\n", "- Oven: 30 minutes.\n", "\n", "Suppose each burner is rated at :math:`1.2 \\kilo\\watt` and the oven at\n", ":math:`1.8 \\kilo\\watt`.\n", "\n", "The total amount of energy expended by the burners is\n", "\n", ".. math::\n", "\n", " w_b &= \\int_{t_0}^t p_b dt\n", " & \\quad \\text{(1.9)}\\\\\n", " &= \\evalAt{p_b t}{0}{(20 + 40 + 15 + 45) \\times 60}\\\\\n", " &= 8640 \\kilo\\joule\\\\\n", " &= 2.4 \\kilo\\watthour\n", "\n", "and the oven is\n", "\n", ".. math::\n", "\n", " w_o &= \\int_{t_0}^t p_o dt\n", " & \\quad \\text{(1.9)}\\\\\n", " &= \\evalAt{p_o t}{0}{30 \\times 60}\\\\\n", " &= 3240 \\kilo\\joule\\\\\n", " &= 0.9 \\kilo\\watthour.\n", "\n", "If electricity costs :math:`12 \\cent\\per\\kilo\\watthour`, the cost of electricity\n", "used in preparing the meal is\n", "\n", ".. math::\n", "\n", " c = 12 \\times (w_b + w_o) = 39.6 \\cent." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.30\n", "=============\n", "\n", "Reliant Energy charges customers as follows:\n", "\n", "- Monthly charge :math:`\\$6`.\n", "- First :math:`250 \\kilo\\watthour @ 2 \\cent\\per\\kilo\\watthour`.\n", "- All additional :math:`\\kilo\\watthour @ 7 \\cent\\per\\kilo\\watthour`.\n", "\n", "If a customer uses :math:`2.436 \\mega\\watthour` in one month, Reliant Energy\n", "will charge\n", "\n", ".. math::\n", "\n", " c = 6 + 0.02 \\times 250 + 0.07 \\times (2436 - 250) = \\$164.02." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.31\n", "=============\n", "\n", "Suppose a :math:`120 \\watt` PC runs for :math:`4 \\hour\\per\\day` while a\n", ":math:`60 \\watt` bulb runs for :math:`8 \\hour\\per\\day`. The energy consumed by\n", "the PC is\n", "\n", ".. math::\n", "\n", " w_{pc} &= \\int_{t_0}^t p_{pc} dt\n", " & \\quad \\text{(1.9)}\\\\\n", " &= \\evalAt{p_{pc} t}{0}{4 \\times 365}\\\\\n", " &= 175200 \\watthour\\\\\n", " &= 175.2 \\kilo\\watthour\n", "\n", "and the bulb is\n", "\n", ".. math::\n", "\n", " w_b &= \\int_{t_0}^t p_b dt\n", " & \\quad \\text{(1.9)}\\\\\n", " &= \\evalAt{p_b t}{0}{8 \\times 365}\\\\\n", " &= 175200 \\watthour\\\\\n", " &= 175.2 \\kilo\\watthour.\n", "\n", "If electricity costs :math:`12 \\cent\\per\\kilo\\watthour`, this household pays per\n", "year\n", "\n", ".. math::\n", "\n", " c = 12 \\times (w_{pc} + w_b) = \\$42.048 \\cent." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Comprehensive Problems\n", "######################\n", "\n", ".. math::\n", "\n", " \\newcommand{\\horsepower}{\\mathrm{hp}}" ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.32\n", "=============\n", "\n", "A telephone wire has a current of :math:`20 \\micro\\ampere` flowing through it.\n", "The time it takes for a charge of :math:`15 \\coulomb` to pass through the wire\n", "is\n", "\n", ".. math::\n", "\n", " Q &= \\int_{t_0}^t i dt\n", " & \\quad \\text{(1.2)}\\\\\n", " 15 &= \\evalAt{it}{0}{t}\\\\\n", " t &= \\frac{15}{i}\\\\\n", " &= 208.\\bar{3} \\hour." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.33\n", "=============\n", "\n", "A lightning bolt carried a current of :math:`2 \\kilo\\ampere` and lasted for\n", ":math:`3 \\milli\\second`. The amount of charge contained in the lightning bolt\n", "is\n", "\n", ".. math::\n", "\n", " Q &= \\int_{t_0}^t i dt\n", " & \\quad \\text{(1.2)}\\\\\n", " &= \\evalAt{it}{0}{0.003}\\\\\n", " &= 6 \\coulomb." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.34\n", "=============\n", "\n", "The power (:math:`\\watt`) consumption of a certain household in one day is given\n", "by\n", "\n", ".. math::\n", "\n", " p(t) = \\begin{cases}\n", " 200 & \\text{if } 0 \\leq t < 6\\\\\n", " 800 & \\text{if } 6 \\leq t < 8\\\\\n", " 200 & \\text{if } 8 \\leq t < 18\\\\\n", " 1200 & \\text{if } 18 \\leq t < 22\\\\\n", " 200 & \\text{if } 18 \\leq t \\leq 24.\n", " \\end{cases}\n", "\n", "(a)\n", "---\n", "\n", "The total energy consumed is\n", "\n", ".. math::\n", "\n", " w &= \\int_{t_0 = 0}^t p dt\n", " & \\quad \\text{(1.9)}\\\\\n", " &= \\int_0^6 200 dt +\n", " \\int_6^8 800 dt +\n", " \\int_8^{18} 200 dt +\n", " \\int_{18}^{22} 1200 dt +\n", " \\int_{22}^{24} 200 dt\\\\\n", " &= 200 (6 - 0) + 800 (8 - 6) + 200 (18 - 8) +\n", " 1200 (22 - 18) + 200 (24 - 22)\\\\\n", " &= 10 \\kilo\\watthour.\n", "\n", "(b)\n", "---\n", "\n", "The average power per hour over the entire day is\n", "\n", ".. math::\n", "\n", " p_{avg} = \\frac{w}{24} = 416.\\bar{6} \\watt." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.35\n", "=============\n", "\n", "The power (:math:`\\mega\\watt`) drawn by an industrial plant between 8:00 and\n", "8:30 A.M. is given by\n", "\n", ".. math::\n", "\n", " p(t) = \\begin{cases}\n", " 5 & \\text{if } 0 \\leq t < 5\\\\\n", " 4 & \\text{if } 5 \\leq t < 10\\\\\n", " 3 & \\text{if } 10 \\leq t < 15\\\\\n", " 8 & \\text{if } 15 \\leq t < 20\\\\\n", " 4 & \\text{if } 20 \\leq t \\leq 30.\n", " \\end{cases}\n", "\n", "The total energy consumed by the plant is\n", "\n", ".. math::\n", "\n", " w &= \\int_{t_0 = 0}^t p dt\n", " & \\quad \\text{(1.9)}\\\\\n", " &= \\int_0^5 5 dt +\n", " \\int_5^{10} 4 dt +\n", " \\int_{10}^{15} 3 dt +\n", " \\int_{15}^{20} 8 dt +\n", " \\int_{20}^{30} 4 dt\\\\\n", " &= \\frac{\n", " 5 (5 - 0) + 4 (10 - 5) + 3 (15 - 10) +\n", " 8 (20 - 15) + 4 (30 - 20)\n", " }{60}\n", " & \\quad \\text{convert } \\minute \\text{ to } \\hour\\\\\n", " &= 2.\\bar{3} \\mega\\watthour." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.36\n", "=============\n", "\n", "A lead-acid battery is rated at :math:`160 \\amperehour`.\n", "\n", "(a)\n", "---\n", "\n", "The maximum current it can supply for :math:`40 \\hour` is\n", "\n", ".. math::\n", "\n", " i &= \\frac{\\Delta q}{\\Delta t}\n", " & \\quad \\text{(1.1)}\\\\\n", " &= \\frac{160}{40}\\\\\n", " &= 4 \\ampere.\n", "\n", "(a)\n", "---\n", "\n", "If it is discharged at :math:`1 \\milli\\ampere`, the battery can last\n", "\n", ".. math::\n", "\n", " \\Delta t &= \\frac{\\Delta q}{i}\n", " & \\quad \\text{(1.1)}\\\\\n", " &= \\frac{160 \\times 3600}{0.001}\\\\\n", " &= 6666.\\bar{6} \\day." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.37\n", "=============\n", "\n", "A :math:`12 \\volt` battery requires a total charge of :math:`40 \\amperehour`\n", "during recharging. The energy supplied to the battery is\n", "\n", ".. math::\n", "\n", " v &= \\frac{\\Delta w}{\\Delta q}\n", " & \\quad \\text{(1.3)}\\\\\n", " \\Delta w &= v \\Delta q\\\\\n", " &= 12 (40 \\times 3600)\\\\\n", " &= 1.728 \\mega\\joule." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.38\n", "=============\n", "\n", "Assuming :math:`1 \\horsepower = 746 \\watt`, a :math:`10 \\horsepower` motor\n", "delivers in 30 minutes\n", "\n", ".. math::\n", "\n", " w &= \\int_{t_0 = 0}^{t} p dt\n", " & \\quad \\text{(1.9)}\\\\\n", " &= \\evalAt{10 \\times 746 t}{0}{30 / 60}\\\\\n", " &= 12 (40 \\times 3600)\\\\\n", " &= 3.73 \\kilo\\watthour." ] }, { "cell_type": "raw", "metadata": { "raw_mimetype": "text/restructuredtext" }, "source": [ "Exercise 1.39\n", "=============\n", "\n", "A :math:`600 \\watt` TV receiver is turned on for :math:`4 \\hour` with nobody\n", "watching it. The total energy expended is\n", "\n", ".. math::\n", "\n", " w &= \\int_{t_0 = 0}^{t} p dt\n", " & \\quad \\text{(1.9)}\\\\\n", " &= \\evalAt{600t}{0}{4}\\\\\n", " &= 2.4 \\kilo\\watthour.\n", "\n", "If electricity costs :math:`10 \\cent\\per\\kilo\\watthour`, the consumer wastes\n", "\n", ".. math::\n", "\n", " c = 10 \\times 2.4 = 24 \\cent." ] } ], "metadata": { "celltoolbar": "Raw Cell Format", "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.7.1" } }, "nbformat": 4, "nbformat_minor": 2 }