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dc_analysis.tex
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3192 lines (2777 loc) · 75.1 KB
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\documentclass[]{beamer}
\usepackage[utf8]{inputenc}
\include{latex-macro}
\usepackage{comment}
\usepackage{icomma}
\usepackage{hyperref}
\usepackage{import}
\usetheme[secheader]{Boadilla}
\setbeamertemplate{navigation symbols}{}
\begin{document}
\title{Circuit analysis: DC Circuits (3 cr)}
\subtitle{Fall 2009 / Class AS09}
\author{Vesa Linja-aho}
\institute{Metropolia}
\frame{
\titlepage
\vfill
\begin{center}
The slides are licensed with CC By 1.0. \url{http://creativecommons.org/licenses/by/1.0/} \\
Slideset version: 1.1
\end{center}
}
\date{\today}
\frame{
\frametitle{Table of Contents}
Click the lecture name to jump onto the first slide of the lecture.
\vfill
\begin{columns}
\begin{column}{0.4\textwidth}
\tableofcontents[sections={1-6}]
\end{column}
\begin{column}{0.4\textwidth}
\tableofcontents[sections={7-12}]
\end{column}
\end{columns}
}
\section{1. lecture}
\frame {
\frametitle{About the Course}
\begin{itemize}
\item Lecturer: M.Sc. Vesa Linja-aho
\item Lectures on Mon 11:00-14:00 and Thu 14:00-16:30, room P113
\item To pass the course: Home assignments and final exam. The exam is on Monday 12th October 2009 at 11:00-14:00.
\item All changes to the schedule are announced in the Tuubi-portal.
\end{itemize}
}
\frame{
\frametitle{The Home Assignments}
\begin{itemize}
\item There are 12 home assignments.
\item Each assignment is graded with 0, 0,5 or 1 points.
\item To pass the course, the student must have at least 4 points from the assignments.
\item Each point exceeding the minimum of 4 points will give you 0,5 extra points in the exam.
\item In the exam, there are 5 assignments, with maximum of 6 points each.
\item To pass the exam, you need to get 15 points from the exam.
\item All other grade limits (for grades 2-5) are flexible.
\end{itemize}
\begin{exampleblock}{Example}
The student has 8 points from the home assignments. He gets 13 points from the exam.
He will pass the exam, because he gets extra points from the home assignments and
his total score is $(8-4)\cdot 0,5+13=15$ points.
\end{exampleblock}
However, is one gets 8 of 12 points from the home assignments, he usually gets more
than 13 points from the exam :-).
}
\frame{
\frametitle{The Course Objectives}
From the curriculum:
\begin{block}{Learning outcomes of the course unit}
\small Basic concepts and basic laws of electrical engineering. Analysis of direct current (DC) circuits.
\end{block}
\begin{block}{Course contents}
\small Basic concepts and basic laws of electrical engineering, analysis methods, controlled sources. Examples and exercises.
\end{block}
}
\frame {
\frametitle{The Course Schedule}
\begin{enumerate}
\item The basic quantities and units. Voltage source and resistance. Kirchhoff's laws and Ohm's law.
\item Conductance. Electric power. Series and parallel circuits. Node. Ground.
\item Current source. Applying the Kirchhoff's laws to solve the circuit. Node-voltage analysis.
\item Exercises on node-voltage analysis.
\item Source transformation.
\item Thévenin equivalent and Norton equivalent.
\item Superposition principle.
\item Voltage divider and current divider.
\item Inductance and capacitance in DC circuits.
\item Controlled sources.
\item Recap.
\item Recap.
\end{enumerate}
}
\frame {
\frametitle{The Course is Solid Ground for Further Studies in Electronics}
The basic knowledge on DC circuits is needed on the courses {\em Circuit Analysis: Basic AC-Theory}, {\em Measuring Technology},
{\em Automotive Electronics 1}, {\em Automotive Electrical Engineering Labs}, $\ldots$
\begin{alertblock}{Important!}
By studying this course well, {\bf studying the upcoming courses will be easier}!
\end{alertblock}
The basics of DC circuits are vital for automotive electronics engineer, just like the
basics of accounting are vital for an auditor, and basics of strength of materials are vital for a bridge-building engineer etc.
}
\frame{
\frametitle{What is Not Covered on This Course}
The basic physical characteristics of electricity is not covered on this course. Questions
like "What is electricity?" are covered on the course {\em Rotational motion and electromagnetism}.
}
\frame{
\frametitle{Studying in Our School}
\begin{itemize}
\item You have an opportunity to learn on the lectures. I can not force you to learn.
\item You have more responsibility on your learning than you had in vocational school or senior high school.
\item 1 cr $\approx$ 26,7 hours of work. 3 cr = 80 hours of work. You will spend 39 hours on the lectures.
\item Which means that you should use about 40 hours of your own time for studying!
\item If I proceed too fast or too slow, please interject me (or tell me by email).
\item Do not hesitate to ask. Ask also the "stupid questions".
\end{itemize}
}
\frame{
\frametitle{What Is Easy and What Is Hard?}
Different things are hard for different people. But my own experience shows that
\begin{itemize}
\item DC analysis is easy, because the math involved is very basic.
\item DC analysis is hard, because the circuits are not as intuitive as, for example, mechanical
systems are.
\end{itemize}
Studying your math courses well is important for the upcoming courses on circuit analysis.
For example, in AC circuits analysis you have to use {\em complex arithmetics}.
}
\frame {
\frametitle{Now, Let's Get into Business}
Any questions on the practical arrangements of the course?
}
\frame {
\frametitle{Electric Current}
\begin{itemize}
\item Electric current is a flow of electric charge.
\item The unit for electric current is the ampere (A).
\item The abbreviation for the quantity is $I$.
\item One may compare the electric current with water flowing in a pipe (so called {\em hydraulic analogy}).
\item The current always circulates in a loop: current does not compress nor vanish.
\item The current in a wire is denoted like this:
\end{itemize}
\begin{center}
\begin{picture}(100,25)(0,0)
\hln{0,0}{100}
\ri{50,0}{I=2\mA}
\end{picture}
\end{center}
}
\frame {
\frametitle{Kirchhoff's Current Law}
\begin{itemize}
\item As mentioned on the previous slide, the current can not vanish anywhere.
\end{itemize}
\begin{block}{Kirchhoff's Current Law (or: Kirchhoff's First Law)}
At any area in an electrical circuit, the sum of currents flowing into that area is equal to the sum of currents flowing out of that area.
\end{block}
\begin{center}
\begin{picture}(100,50)(0,0)
\hln{0,0}{100}
\hln{0,50}{100}
\vln{0,0}{50}
\vln{50,0}{50}
\vln{100,0}{50}
\ri{25,0}{I_1=3\mA}
\ri{75,0}{I_2=2\mA}
\ui{50,25}{I_3=1\mA}
\end{picture}
\end{center}
If you draw a circle in any place in the circuit, you can observe that there is as the same amount of current flowing into
the circle and out from the circle!
}
\frame {
\frametitle{Be Careful with Signs}
\begin{itemize}
\item One can say: "The balance of my account -50 euros" or equally "I owe 50 euros to my bank".
\item One can say: "The profit of the company was -500000 euros" or equally "The loss of the company was 500000 euros".
\item If you measure a current with an ammeter and it reads $-15 \mA$, by reversing the wires of the ammeter it
will show $15 \mA$.
\item The sign of the current shows the direction of the current. The two circuits below are exactly identical.
\end{itemize}
\begin{picture}(100,50)(0,0)
\hln{0,0}{100}
\hln{0,50}{100}
\vln{0,0}{50}
\vln{50,0}{50}
\vln{100,0}{50}
\ri{25,0}{I_1=3\mA}
\ri{75,0}{I_2=2\mA}
\ui{50,25}{I_3=1\mA}
\end{picture}
\begin{picture}(100,50)(-50,0)
\hln{0,0}{100}
\hln{0,50}{100}
\vln{0,0}{50}
\vln{50,0}{50}
\vln{100,0}{50}
\li{25,0}{I_{\rm a}=-3\mA}
\li{75,0}{I_{\rm b}=-2\mA\hspace{-1cm}}
\ui{50,25}{I_3=1\mA}
\end{picture}
}
\frame {
\frametitle{Voltage}
\begin{itemize}
\item The potential difference between two points is called voltage.
\item The abbreviation for the quantity is $U$.
\item In circuit theory, it is insignificant how the potential difference is generated (chemically, by induction etc.).
\item The unit of voltage is the volt (V).
\item One may compare the voltage with a pressure difference in hydraulic system, or to a difference in altitude.
\item Voltage is denoted with an arrow between two points.
\end{itemize}
\begin{center}
\begin{picture}(50,50)(0,0)
\vst{0,0}{12 \V}
\du{20,0}{U=12 \V}
\hln{0,0}{20}
\hln{0,50}{20}
\end{picture}
\end{center}
}
\frame {
\frametitle{Kirchhoff's voltage law}
\begin{itemize}
\item The voltage between two points is the same, regardless of the path chosen.
\item This is easy to understand by using the analogy of differences in altitude. If you
leave your home, go somewhere and return to your home, you have traveled uphill as much
as you have traveled downhill.
\end{itemize}
\begin{block}{Kirchhoff's Voltage Law (or: Kirchhoff's Second Law)}
The directed {\bf sum of the voltages around any closed circuit is zero}.
\end{block}
\begin{center}
\begin{picture}(100,50)(0,0)
\hst{0,0}{1,5 \V}
\hst{50,0}{1,5 \V}
\hst{100,0}{1,5 \V}
\vln{0,0}{50}
\vln{150,0}{50}
\hln{0,50}{50}
\hln{100,50}{50}
\lu{50,50}{4,5 \V}
\cn{50,50}
\cn{100,50}
\end{picture}
\end{center}
}
\frame{
\frametitle{Ohm's law}
\begin{itemize}
\item {\em Resistance } is a measure of the degree to which an object opposes an electric current through it.
\item The larger the current, the larger the voltage -- and vice versa.
\item The abbreviation of the quantity is $R$ and the unit is ($\ohm$) (ohm).
\item The definition of resistance is the ratio of the voltage over the element divided with
the current through the element. $R=U/I$
\end{itemize}
\begin{center}
$U=RI$
\begin{picture}(50,50)(0,0)
\hz{0,0}{R}
\ru{0,10}{U}
\ri{5,0}{I}
\hln{50,0}{15}
\hln{-15,0}{15}
\end{picture}
\end{center}
}
\frame {
\frametitle{Definitions}
\begin{description}
\item[Electric circuit] A system consisting of compontents, in where electric current flows.
\item[Direct current (DC)] The electrical quantities (voltage and current) are constant (or nearly constant) over time.
\item[Direct current circuit] An electric circuit, where voltages and currents are constant over time.
\end{description}
\begin{exampleblock}{Example}
In a flashlight, there is a direct current circuit consisting of a battery/batteries, a switch and a bulb.
In a bicycle there is an alternating current circuit (dynamo and bulb).
\end{exampleblock}
}
\frame {
\frametitle{An alternate definition for direct current}
One may define also that direct current means a current, which does not change its direction (sign),
but the magnitude of the current can vary over time. For example, a simple lead acid battery charger
outputs a pulsating voltage, which varies between 0 V ... $\approx$ 18 V. This can be also called DC voltage.
\begin{alertblock}{Agreement}
On this course, we define DC to mean constant voltage and current. The magnitude and sign
are constant over time.
\end{alertblock}
}
\frame{
\frametitle{A simple DC circuit}
\begin{itemize}
\item A light bulb is wired to a battery. The resistance of the filament is $10 \ohm$.
\end{itemize}
\begin{center}
\begin{picture}(50,50)(0,0)
\vst{0,0}{12 \V}
\uncover<-1>{\vlamp{50,0}{}}
\uncover<2->{\vz{50,0}{10\ohm}}
\uncover<-3>{\ri{25,50}{I=?}}
\uncover<4->{\ri{25,50}{I=1,2\A}}
\hln{0,0}{50}
\hln{0,50}{50}
\uncover<3->{\du{65,0}{12\V}}
\end{picture}
\end{center}
\uncover<4->{$U=RI$\\
$I=\frac{U}{R}=\frac{12 \V}{10 \ohm}=1,2 \A$}
}
\begin{comment}
\frame{
\frametitle{Oppikirja}
Tällä luennolla käsiteltiin kirjan Kimmo Silvonen: {\em Sähkötekniikka ja elektroniikka} kappaleet:
\begin{description}
\item[1.1.1] Sähkövirta ja Kirchhoffin virtalaki
\item[1.1.3] Potentiaaliero ja Kirchhoffin jännitelaki
\item[1.2.1] Ohmin laki
\end{description}
Koska sivunumerointi saattaa vaihdella painoksittain, viittaan kappaleen numeroihin.
}
\end{comment}
\frame{
\frametitle{Homework 1 (released 31st Aug, to be returned 3rd Sep)}
\begin{itemize}
\item The homework are to be returned at the beginning of the next lecture.
\item Remember to include your name and student number.
\end{itemize}
\begin{block}{Homework 1}
Find the current $I$.
\end{block}
\begin{center}
\begin{picture}(50,100)(0,0)
\vst{0,0}{1,5 \V}
\vst{0,50}{1,5 \V}
\vz{50,25}{R=20\ohm\hspace{-2.5cm}}
\vln{50,0}{25}
\vln{50,75}{25}
\hln{0,100}{50}
\hln{0,0}{50}
\di{50,25}{I}
\end{picture}
\end{center}
}
\section{2. lecture}
\frame{
\frametitle{Homework 1 - Model solution}
\begin{block}{Homework 1}
Find the current $I$.
\end{block}
\begin{center}
\begin{picture}(50,100)(0,0)
\vst{0,0}{1,5 \V}
\vst{0,50}{1,5 \V}
\vz{50,25}{R=20\ohm\hspace{-2.5cm}}
\vln{50,0}{25}
\vln{50,75}{25}
\hln{0,100}{50}
\hln{0,0}{50}
\di{50,25}{I}
\uncover<2->{\du{12,0}{U_1} }
\uncover<2->{\du{12,50}{U_2} }
\uncover<2->{\du{41,25}{\hspace{-0.65cm}U_{\rm R}} }
\end{picture}
\end{center}%\uncover<3->{\[ U_1+U_2-U_{\rm R}=0 \Leftrightarrow U_{\rm R}=U_1+U_2 \]}\uncover<4->{\[ U=RI \Rightarrow U_{\rm R}=RI \Rightarrow I=\frac{U_{\rm R}}{R}=\frac{U_1+U_2}{R}=\frac{1,5 \V+1,5 \V}{20\ohm}=150 \mA \]}
\uncover<2->{\[ U_1+U_2-U_{\rm R}=0 \Leftrightarrow U_{\rm R}=U_1+U_2 \]}
\uncover<3->{\vspace{-\baselineskip}
\[ U=RI \Rightarrow U_{\rm R}=RI \Rightarrow I=\frac{U_{\rm R}}{R}=\frac{U_1+U_2}{R}=\frac{1,5 \V+1,5 \V}{20\ohm}=150 \mA \]}
}
\frame{
\frametitle{Conductance}
\begin{itemize}
\item Resistance is a measure of the degree to which an object opposes an electric current through it.
\item The inverse of resistance is {\bf conductance}. The symbol for conductance is
$G$ and the unit is Siemens (S).
\item Conductance measures how easily electricity flows along certain element.
\item For example, if resistance $R=10 \ohm$ then conductance $G=0,1\Siemens$.
\end{itemize}
\begin{center}
$G=\frac{1}{R} \qquad U=RI \Leftrightarrow GU=I$
\begin{picture}(50,50)(0,0)
\hz{0,0}{G=\frac{1}{R}}
\ru{0,10}{U}
\ri{5,0}{I}
\hln{50,0}{15}
\hln{-15,0}{15}
\end{picture}
\end{center}
}
\frame {
\frametitle{Electric Power}
\begin{itemize}
\item In physics, power is the rate at which work is performed.
\item The symbol for power is $P$ and the unit is the Watt (W).
\item The DC power consumed by an electric element is $P=UI$\\[1cm] \begin{picture}(50,10)(-30,-5) \hz{0,0}{} \ri{8,0}{I} \ru{0,10}{U} \end{picture}
\item If the formula outputs a positive power, the element is consuming power from the circuit. If the formula outputs a negative power, the element is delivering power to the circuit.
\end{itemize}
}
\frame{
\frametitle{Electric Power}
\begin{block}{Energy can not be created nor destroyed}
The power consumed by the elements in the circuit = the power delivered by the elements in the circuit.
\end{block}
\begin{center}
\begin{picture}(200,50)(0,0)
\vst{0,0}{E}
\vz{100,0}{R}
\hln{0,0}{100}
\hln{0,50}{100}
\ui{0,45}{I}
\di{100,42}{I}
\txt{200,50}{I=\frac{U}{R}}
\txt{200,25}{P_R=UI=U\frac{U}{R}=\frac{U^2}{R}}
\txt{200,0}{P_E=U\cdot(-I)=U\frac{-U}{R}=-\frac{U^2}{R}}
\end{picture}
\end{center}
The power delivered by the voltage source is consumed by the resistor.
}
\frame {
\frametitle{Series and Parallel Circuits}
\begin{alertblock}{Definition: series circuit}
The elements are in series, if they are connected so that the same current flows through the elements.
\end{alertblock}
\begin{alertblock}{Definition: parallel circuit}
The elements are in parallel, if they are connected so that there is the same voltage across them.
\end{alertblock}
}
\frame {
\frametitle{Series and Parallel Circuits}
\begin{exampleblock}{Series circuit}
\begin{center}
\begin{picture}(100,20)(0,-10)
\hz{0,0}{}
\hz{50,0}{}
\ri{-10,0}{I}
\ri{110,0}{I}
\hln{-50,0}{50}
\hln{100,0}{50}
\end{picture}
\end{center}
\end{exampleblock}
\begin{exampleblock}{Parallel Circuit}
\begin{center}
\begin{picture}(50,75)(0,-5)
\hz{0,0}{}
\hz{0,50}{}
\vln{0,0}{50}
\vln{50,0}{50}
\ru{0,10}{U}
\ru{0,60}{U}
\end{picture}
\end{center}
\end{exampleblock}
}
\frame {
\frametitle{Resistors in series and in parallel}
\begin{exampleblock}{In series}
\begin{center}
\begin{picture}(100,20)(50,-15)
\hz{0,0}{R_1}
\hz{50,0}{R_2}
\txt{125,0}{\Longleftrightarrow}
\hz{150,0}{R=R_1+R_2}
\end{picture}
\end{center}
\end{exampleblock}
\begin{exampleblock}{In parallel}
\begin{center}
\begin{picture}(100,75)(0,-20)
\hz{0,0}{R_1}
\hz{0,50}{R_2}
\hln{-25,25}{25}
\hln{50,25}{25}
\vln{0,0}{50}
\vln{50,0}{50}
\txt{110,25}{\Longleftrightarrow}
\hz{130,25}{\vspace{-1cm}R=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}}}
\end{picture}
\end{center}
\end{exampleblock}
Or, by using conductances: $G=G_1+G_2$.
}
\frame{
\frametitle{Resistors in series and in parallel}
\begin{itemize}
\item The formulae on the previous slide can be applied to an arbitrary number of resistors. For instance,
the total resistance of five resistors in series is $R=R_1+R_2+R_3+R_4+R_5$.
\end{itemize}
}
\frame{
\frametitle{Voltage Sources in Series}
\begin{itemize}
\item The voltages can be summed like resistances, but be careful with correct signs.
\item Voltage sources in parallel are inadmissible in circuit theory. There can not be two different
voltages between two nodes at the same time.
\end{itemize}
\begin{center}
\begin{picture}(100,25)(0,-30)
\hst{0,0}{E_1}
\hlst{50,0}{E_2}
\hst{100,0}{E_3}
\cn{0,0}
\cn{150,0}
\end{picture}
\begin{picture}(100,25)(0,0)
\txt{75,23}{\Longleftrightarrow}
\hst{50,0}{E=E_1-E_2+E_3}
\cn{50,0}
\cn{100,0}
\end{picture}
\end{center}
}
\frame{
\frametitle{What Series and Parallel Circuits are NOT}
\begin{itemize}
\item Just the fact that two components seem to be one after the other, does not mean that they are in series.
\item Just the fact that two components seem to be side by side, does not mean that they are in parallel.
\item In the figure below, which of the resistors are in parallel and which are in series with each other?
\end{itemize}
\begin{center}
\begin{picture}(100,50)(0,0)
\vst{0,0}{E_1}
\vst{100,0}{E_2}
\vz{50,0}{R_3}
\hz{0,50}{R_1}
\hz{50,50}{R_2}
\hln{0,0}{100}
\end{picture}
\end{center}
\pause
\vspace{-0.2cm}
\begin{alertblock}{Solution}
\scriptsize
None! $E_1$ ja $R_1$ are in series and $E_2$ ja $R_2$ are in series. Both of these serial circuits are in parallel with $R_3$. But no two resistors are in parallel nor in series.
\end{alertblock}
}
\frame{
\frametitle{Terminal and Gate}
\begin{itemize}
\item A point which provides a point of connection to external circuits is called a terminal (or pole).
\item Two terminals form a gate.
\item An easy example: a car battery with internal resistance.
\end{itemize}
\begin{center}
\begin{picture}(100,50)(0,0)
\vst{0,0}{E}
\hz{0,50}{R_{\rm S}}
\hln{0,0}{50}
\out{50,0}
\out{50,50}
\end{picture}
\end{center}
}
\frame{
\frametitle{Node}
\begin{itemize}
\item {\bf A node} means an area in the circuit where there are no potential differences, or alternatively a place where two or more circuit elements meet.
\item A "for dummies" --way to find nodes in the circuit: put your pen on a wire in the circuit. Start coloring
the wire, and backtrack when your pen meets a circuit element. The area you colored is one node.
\item How many nodes are there in the circuit below?
\end{itemize}
\begin{center}
\begin{picture}(150,50)(0,0)
\vst{0,0}{E}
\hz{0,50}{R_1}
\hz{50,50}{R_3}
\hz{100,50}{R_5}
\vz{50,0}{R_2}
\vz{100,0}{R_4}
\vz{150,0}{R_6}
\hln{0,0}{150}
\ri{8,50}{I}
\end{picture}
%$R_1=R_2=R_3=R_4=R_5=R_6=1\ohm\qquad E=9\V$
\end{center}
}
\frame{
\frametitle{Ground}
\begin{itemize}
\item One of the nodes in the circuit can be appointed the ground node.
\item By selecting one of the nodes to be the ground node, the circuit diagram usually appear cleaner.
\item The car battery is connected to the chassis of the car. Therefore it is convenient to handle the chassis
as the ground node.
\item When we say "the voltage of this node is 12 volts" it means that the voltage between that node
and the ground node is 12 volts.
\end{itemize}
\begin{center}
\begin{picture}(150,50)(0,0)
\vst{0,0}{E}
\hz{0,50}{R_1}
\hz{50,50}{R_3}
\hz{100,50}{R_5}
\vz{50,0}{R_2}
\vz{100,0}{R_4}
\vz{150,0}{R_6}
\hln{0,0}{150}
\ri{8,50}{I}
\hgp{0,0}
\cn{0,0}
\end{picture}
\end{center}
}
\frame{
\frametitle{Ground}
\begin{itemize}
\item The ground node can be connected to the chassis of the device or it can be leave not connected to the chassis.
\item Therefore, the existence of the ground node does not mean that the device is "grounded".
\item The circuit on the previous slide can be presented also like this:
\end{itemize}
\begin{center}
\begin{picture}(150,50)(0,0)
\vst{0,0}{E}
\hz{0,50}{R_1}
\hz{50,50}{R_3}
\hz{100,50}{R_5}
\vz{50,0}{R_2}
\vz{100,0}{R_4}
\vz{150,0}{R_6}
%\hln{0,0}{150}
\ri{8,50}{I}
\hgp{0,0}
\hgp{50,0}
\hgp{100,0}
\hgp{150,0}
%\cn{0,0}
\end{picture}
\end{center}
}
%\subsection{Viitteet oppikirjaan ja kotitehtävä}
\begin{comment}
\frame{
\frametitle{Oppikirja}
Tällä luennolla käsiteltiin kirjan Kimmo Silvonen: {\em Sähkötekniikka ja elektroniikka} kappaleet:
\begin{description}
\item[1.2.2] [Siemensin laki ja] konduktanssi eli johtokyky
\item[1.3.1] Tehon ja energiankulutuksen laskeminen
\item[1.4.1] Sarjaankytkentä
\item[1.4.2] Rinnankytkentä
\item[1.5.1] Vastusten sarjaankytkentä
\item[1.5.2] Vastusten rinnankytkentä
\item[1.4.5] Napa, portti, maa
\end{description}
Koska sivunumerointi saattaa vaihdella painoksittain, viittaan kappaleen numeroihin.
}
\end{comment}
\frame{
\frametitle{Homework 2 (released 3rd Sep, to be returned 7th Sep)}
\begin{block}{Homework 2}
Find the current $I$.
\end{block}
\begin{center}
\begin{picture}(150,50)(0,0)
\vst{0,0}{E}
\hz{0,50}{R_1}
\hz{50,50}{R_3}
\hz{100,50}{R_5}
\vz{50,0}{R_2}
\vz{100,0}{R_4}
\vz{150,0}{R_6}
\hln{0,0}{150}
\ri{8,50}{I}
\end{picture}
$R_1=R_2=R_3=R_4=R_5=R_6=1\ohm\qquad E=9\V$
\end{center}
}
\section{3. lecture}
\frame{
\frametitle{Homework 2 - Model solution}
\begin{block}{Homework 2}
Find the current $I$.
\end{block}
\begin{center}
\begin{picture}(150,50)(0,0)
\vst{0,0}{E}
\hz{0,50}{R_1}
\hz{50,50}{R_3}
\hz{100,50}{R_5}
\vz{50,0}{R_2}
\vz{100,0}{R_4}
\vz{150,0}{R_6}
\hln{0,0}{150}
\ri{8,50}{I}
\end{picture}
$R_1=R_2=R_3=R_4=R_5=R_6=1\ohm\qquad E=9\V$
\end{center}
\begin{itemize}
\item $R_5$ ja $R_6$ are in series. The total resistance of the serial connection is $R_5+R_6=2\ohm$.
\item Furthermore, the serial connection is in parallel with $R_4$. The resistance of this parallel circuit is
$\frac{1}{\frac{1}{1}+\frac{1}{2}}\ohm=\frac{2}{3}\ohm$.
\end{itemize}
}
\frame{
\frametitle{Solution continues}
\begin{itemize}
\item $R_3$ is in series with the parallel circuit calculated on the previous slide. The resistance for this circuit is $R_3+\frac{2}{3}\ohm
=\frac{5}{3}\ohm$.
\item And the serial connection is in parallel with $R_2$. The resistance for the parallel circuit is
$\frac{1}{(\frac{5}{3})^{-1}+\frac{1}{1}}=\frac{5}{8}\ohm$.
\item Lastly, $R_1$ is in series with the resistance computed in the previous step. Therefore, the total resistance seen by voltage source $E$ is
$\frac{5}{8}\ohm + R_1=\frac{13}{8}\ohm$.
\item The current $I$ is computed from Ohm's law $I=\frac{E}{\frac{13}{8}\ohm}=\frac{72}{13}\A\approx5,5\A$.
\end{itemize}
}
\frame{
\frametitle{The Current Source}
\begin{itemize}
\item The current source is a circuit element which delivers a certain current throught it, just like the voltage source
keeps a certain voltage between its nodes.
\item The current can be constant or it can vary by some rule.
\end{itemize}
\begin{center}
\begin{picture}(100,50)(0,0)
\vj{0,0}{J}
\vz{50,0}{R}
\hln{0,0}{50}
\hln{0,50}{50}
\end{picture}
\end{center}
}
\frame{
\frametitle{The Current Source}
\begin{itemize}
\item If there is a current source in a wire, you know the current of that wire.
\end{itemize}
\begin{center}
\begin{picture}(100,50)(0,0)
\vj{0,0}{J=1\A}
\vz{50,0}{R_1}
\hln{0,0}{100}
\hln{0,50}{100}
\ri{25,50}{I=1\A}
\vz{100,0}{R_2}
\end{picture}
\end{center}
}
\frame{
\frametitle{Applying Kirchhoff's Laws Systematically to the Circuit}
When solving a circuit, it is highly recommended to use a systematic mehtod to find the voltages and/or currents.
Otherwise it is easy to end up with writing a bunch of equations which can not be solved.
One systematic method is called the {\bf nodal analysis}:
\begin{enumerate}
\item Name each current in the circuit.
\item Select one node as the ground node. Assign a variable for each voltage between each node and ground node.
\item Write an equation based on Kirchhoff's current law for each node (except the ground node).
\item State the voltage of each resistor by using the node voltage variables in step 2. Draw the voltage arrows
at the same direction you used for the current arrows (this makes it easier to avoid sign mistakes).
\item State every current by using the voltages and substitute them into the current equations in step 2.
\item Solve the set of equations to find the voltage(s) asked.
\item If desired, solve the currents by using the voltages you solved.
\end{enumerate}
}
\frame{
\frametitle{Example}
Find the current $I$.
\begin{center}
\begin{picture}(150,100)(0,0)
\vst{0,0}{E_1}
\vst{100,0}{E_2}
\vz{50,0}{R_3\hspace{-0.65cm}}
\hz{0,50}{R_1\vspace{0.8cm}}
\hz{50,50}{R_2\vspace{0.8cm}}
\di{50,1}{I}
\hln{0,0}{100}
\pause
\color{red}
\ri{48,50}{I_1}
\li{52,50}{}
\txt{55,60}{I_2}
\pause
\color{blue}
\hgp{25,0}
\du{58,0}{U_3}
\pause
\color{violet}
\txt{160,25}{I=I_1+I_2}
\pause
\color{cyan}
\ru{0,60}{E_1-U_3}
\lu{50,60}{E_2-U_3}
\end{picture}
\end{center}
\pause
\[
\color{magenta} \frac{U_3}{R_3}=\frac{E_1-U_3}{R_1}+\frac{E_2-U_3}{R_2}
\pause
\color{orange} \Longrightarrow
U_3=R_3\frac{R_2E_1+R_1E_2}{R_1R_2+R_2R_3+R_1R_3}
\]
\pause
\[
\color{brown} I=\frac{U_3}{R_3}=\frac{R_2E_1+R_1E_2}{R_1R_2+R_2R_3+R_1R_3}
\]
}
\frame{
\frametitle{Some Remarks}
\begin{itemize}
\item There are many methods for writing the circuit equations, and there is no such thing as "right" method.
\item The only requirement is that you follow Kirchhoff's laws and Ohm's law\footnote{Ohm's law can only be utilized for
resistors. If you have other elements, you must know their current-voltage equation.}
and you have an equal
number of equations and unknowns.
\item If there is a current source in the circuit, it will (usually) make the circuit easier to solve, as
you then have one unknown less to solve.
\item By using conductances instead of resistances, the equations look a little cleaner.
\end{itemize}
}
\frame{
\frametitle{Another Example}
\begin{center}
\begin{picture}(150,100)(0,0)
\vst{0,0}{E_1}
\vst{150,0}{E_2}
\vz{50,0}{R_3\hspace{-0.65cm}}
\hz{0,50}{R_1\vspace{0.8cm}}
\hz{50,50}{R_2\vspace{0.8cm}}
\hln{0,0}{150}
\vz{100,0}{R_4\hspace{-0.65cm}}
\hz{100,50}{R_5\vspace{0.8cm}}
\color{green}
\ri{8,50}{I_1}
\ri{58,50}{I_2}
\di{50,1}{I_3}
\ri{108,50}{I_5}
\di{100,1}{I_4}
\txt{200,25}{I_1=I_2+I_3}
\txt{200,10}{I_2=I_4+I_5}
\color{blue}
\du{58,0}{U_3}
\du{108,0}{U_4}
\end{picture}
\end{center}
\[\color{blue}
\frac{E_1-U_3}{R_1}=\frac{U_3-U_4}{R_2}+\frac{U_3}{R_3} \quad \mbox{ja}\quad
\frac{U_3-U_4}{R_2}=\frac{U_4}{R_4}+\frac{U_4-E_2}{R_5}
\]
\[\color{red}
G_1(E_1-U_3)=G_2(U_3-U_4)+G_3U_3\ \mbox{ja}\
G_2(U_3-U_4)=G_4U_4+G_5(U_4-E_2)
\]
Two equations, two unknowns $\to$ can be solved. Use conductances!
}
\frame{
\frametitle{Some Remarks}
\begin{itemize}
\item There are many other methods available too: mesh analysis, modified nodal analysis, branch current method \ldots
\item If there are ideal voltage sources in the circuit (=voltage sources which are connected to a node without a series resistance),
you need one more unknown (the current of the voltage source) and one more equation (the voltage source will determine the voltage between
the nodes it is connected to).
\end{itemize}
}
\frame{
\frametitle{Homework 3 (released 7th Sep, to be returned 10th Sep)}
\begin{block}{Homework 3a)}
Find the current $I_4$.
\end{block}
\begin{block}{Homework 3b)}
Verify your solution by writing down all the voltages and currents to the circuit diagram and checking that
the solution does not contradict Ohm's and Kirchhoff's laws.
\end{block}
\begin{center}
\begin{picture}(150,50)(0,0)