<?xml version="1.0" encoding="utf-8"?>
<document xmlns="http://cnx.rice.edu/cnxml" xmlns:m="http://www.w3.org/1998/Math/MathML" xmlns:md="http://cnx.rice.edu/mdml/0.4" xmlns:bib="http://bibtexml.sf.net/" xmlns:q="http://cnx.rice.edu/qml/1.0" id="m0014" module-id="" cnxml-version="0.6">
  <title>Electric Circuits and Interconnection Laws</title>
  
  <metadata xmlns:md="http://cnx.rice.edu/mdml/0.4">
  <!-- WARNING! The 'metadata' section is read only. Do not edit below.
       Changes to the metadata section in the source will not be saved. -->
  <md:content-id>m0014</md:content-id>
  <md:title>Electric Circuits and Interconnection Laws</md:title>
  <md:version>2.27</md:version>
  <md:created>2000/07/06</md:created>
  <md:revised>2009/06/04 11:33:20.843 GMT-5</md:revised>
  <md:authorlist>
    <md:author id="dhj">
        <md:firstname>Don</md:firstname>
        <md:surname>Johnson</md:surname>
        <md:fullname>Don Johnson</md:fullname>
        <md:email>dhj@rice.edu</md:email>
    </md:author>
  </md:authorlist>
  <md:maintainerlist>
    <md:maintainer id="dhj">
        <md:firstname>Don</md:firstname>
        <md:surname>Johnson</md:surname>
        <md:fullname>Don Johnson</md:fullname>
        <md:email>dhj@rice.edu</md:email>
    </md:maintainer>
    <md:maintainer id="rha">
        <md:firstname>Roy</md:firstname>
        <md:surname>Ha</md:surname>
        <md:fullname>Roy Ha</md:fullname>
        <md:email>rha@alumni.rice.edu</md:email>
    </md:maintainer>
    <md:maintainer id="bfite">
        <md:firstname>Benjamin</md:firstname>
        <md:surname>Fite</md:surname>
        <md:fullname>Benjamin Fite</md:fullname>
        <md:email>bfite@rice.edu</md:email>
    </md:maintainer>
  </md:maintainerlist>
  <md:license href="http://creativecommons.org/licenses/by/1.0"/>
  <md:licensorlist>
    <md:licensor id="dhj">
        <md:firstname>Don</md:firstname>
        <md:surname>Johnson</md:surname>
        <md:fullname>Don Johnson</md:fullname>
        <md:email>dhj@rice.edu</md:email>
    </md:licensor>
  </md:licensorlist>
  <md:keywordlist>
    <md:keyword>circuit</md:keyword>
    <md:keyword>Kirchoff</md:keyword>
    <md:keyword>node</md:keyword>
  </md:keywordlist>
  <md:subjectlist>
    <md:subject>Science and Technology</md:subject>
  </md:subjectlist>
  <md:abstract>A circuit is the electrical/physical analogue of a system.  Kirchoff's laws provde
connection rules about "nodes" in a circuit to determine relations between various
circuit elements.
</md:abstract>
  <md:language>en</md:language>
  <!-- WARNING! The 'metadata' section is read only. Do not edit above.
       Changes to the metadata section in the source will not be saved. -->
</metadata>

<content>
    <para id="para1">
      A <term>circuit</term> connects circuit elements together in a
      specific configuration designed to transform the source signal
      (originating from a voltage or current source) into another
      signal—the output—that corresponds to the current
      or voltage defined for a particular circuit element. A simple
      resistive circuit is shown in <link target-id="simplecircuit" strength="2"/>.  This circuit is the electrical embodiment of a
      system having its input provided by a source system producing
      <m:math>
	<m:apply>
	  <m:ci type="fn">
	    <m:msub>
	      <m:mi>v</m:mi>
	      <m:mtext>in</m:mtext>
	    </m:msub>
	  </m:ci>
	  <m:ci>t</m:ci>
	</m:apply>
      </m:math>.
    </para>

    <figure id="simplecircuit" orient="vertical">
      <subfigure id="simplecircuit1">                                    
	<media id="id1165859867402" alt="">
          <image src="circuit4.png" mime-type="image/png"/>
          <image src="circuit4.eps" mime-type="application/postscript"/>
        </media>
      </subfigure>
      <subfigure id="simplecircuit2">
	<media id="id6898209" alt="">
          <image src="circuit4a.png" mime-type="image/png"/>
          <image src="circuit4a.eps" mime-type="application/postscript"/>
        </media>
      </subfigure>
      <subfigure id="id6739830">
	<media id="id6716773" alt="">
          <image src="system4.png" mime-type="image/png"/>
          <image src="system4.eps" mime-type="application/postscript"/>
        </media>
      </subfigure> 
      <caption>
	The circuit shown in the top two figures is perhaps the
	simplest circuit that performs a signal processing
	function. On the bottom is the block diagram that corresponds
	to the circuit. The input is provided by the voltage source
        <m:math>
          <m:ci>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mtext>in</m:mtext>
            </m:msub>
          </m:ci>
        </m:math>
	and the output is the voltage    
	<m:math>
	  <m:ci>
	    <m:msub>
	      <m:mi>v</m:mi>
	      <m:mtext>out</m:mtext>
	    </m:msub>
	  </m:ci>
	</m:math>
	across the resistor label 
	<m:math>
	  <m:ci>
	    <m:msub>
	      <m:mi>R</m:mi>
	      <m:mn>2</m:mn>
	    </m:msub>
	  </m:ci>
	</m:math>.
	As shown in the middle, we <emphasis>analyze</emphasis> the
	circuit—understand what it accomplishes—by
	defining currents and voltages for all circuit elements, and
	then solving the circuit and element equations.
      </caption>    
    </figure>
    
    <para id="para2">To understand what this circuit accomplishes, we want to
      determine the voltage across the resistor labeled by its value 
      <m:math>
	<m:ci>
	  <m:msub>
	    <m:mi>R</m:mi>
	    <m:mn>2</m:mn>
	  </m:msub>
	</m:ci>
      </m:math>. 
      Recasting this problem mathematically, we need to solve some set
      of equations so that we relate the output voltage
      <m:math>
	<m:ci>
	  <m:msub>
	    <m:mi>v</m:mi>
	    <m:mtext>out</m:mtext>
	  </m:msub>
	</m:ci>
      </m:math>
      to the source voltage. It would be simple—a little too
      simple at this point—if we could instantly write down the
      one equation that relates these two voltages. Until we have more
      knowledge about how circuits work, we must write a set of
      equations that allow us to find <emphasis>all</emphasis> the
      voltages and currents that can be defined for every circuit
      element. Because we have a three-element circuit, we have a
      total of six voltages and currents that must be either specified
      or determined.  You can define the directions for current flow
      and positive voltage drop <emphasis>any way you
      like</emphasis>. When two people solve a circuit their own ways,
      the signs of their variables may not agree, but current flow and
      voltage drop values for each element will agree.  Do recall in <link document="m0012" strength="3">defining your voltage and current
      variables</link> that the <emphasis>v-i</emphasis> relations for
      the elements presume that positive current flow is in the same
      direction as positive voltage drop. Once you define voltages and
      currents, we need six nonredundant equations to solve for the
      six unknown voltages and currents. By specifying the source, we
      have one; this amounts to providing the source's
      <emphasis>v-i</emphasis> relation. The <emphasis>v-i</emphasis>
      relations for the resistors give us two more. We are only
      halfway there;
      where do we get the other three equations we need?
    </para>
    
    <para id="para3">What we need to solve every circuit problem are
      mathematical statements that express how the circuit elements are interconnected. Said another way, we need the laws that govern the
      electrical connection of circuit elements. First of all, the
      places where circuit elements attach to each other are called
      <term>nodes</term>. Two nodes are explicitly indicated in <link target-id="simplecircuit" strength="2"/>; a third is at the bottom
      where the voltage source and resistor
      <m:math>
	<m:ci>
	  <m:msub>
	    <m:mi>R</m:mi>
	    <m:mn>2</m:mn>
	  </m:msub>
	</m:ci>
      </m:math> are connected. Electrical engineers tend to draw
      circuit diagrams—schematics— in a rectilinear
      fashion. Thus the long line connecting the bottom of the voltage
      source with the bottom of the resistor is intended to make the
      diagram look pretty. This line simply means that the two
      elements are connected together.  <term>Kirchoff's Laws</term>,
      one for <link target-id="voltage" strength="2">voltage</link> and
      one for <link target-id="current" strength="2">current</link>,
      determine what a connection among circuit elements means.  These
      laws can help us analyze this circuit.
    </para>

    <section id="current">
      <title>Kirchoff's Current Law</title>
      <para id="para1wra">
	At every node, the sum of all currents entering a node must
	equal zero. What this law means physically is that charge
	cannot accumulate in a node; what goes in must come out. In
	the example, <link target-id="simplecircuit" strength="2"/>,
	below we have a three-node circuit and thus have three KCL
	equations.

	  <m:math display="block">
	    <m:mtable columnalign="right">
	      <m:mtr>
	        <m:mtd>
	          <m:apply><m:eq/>
                <m:apply><m:minus/>
		          <m:apply><m:minus/><m:ci>i</m:ci></m:apply>
		          <m:ci><m:msub><m:mi>i</m:mi><m:mn>1</m:mn></m:msub></m:ci>
                </m:apply>
                <m:cn>0</m:cn>
	          </m:apply>
	        </m:mtd>
	      </m:mtr>

	      <m:mtr>
	        <m:mtd>
	          <m:apply><m:eq/>
                <m:apply><m:minus/>
		          <m:ci><m:msub><m:mi>i</m:mi><m:mn>1</m:mn></m:msub></m:ci>
		          <m:ci><m:msub><m:mi>i</m:mi><m:mn>2</m:mn></m:msub></m:ci>
                </m:apply>
                <m:cn>0</m:cn>
	          </m:apply>
	        </m:mtd>
	      </m:mtr>

	      <m:mtr>
	        <m:mtd>
	          <m:apply><m:eq/>
                <m:apply><m:plus/>
		          <m:ci>i</m:ci>
		          <m:ci><m:msub><m:mi>i</m:mi><m:mn>2</m:mn></m:msub></m:ci>
                </m:apply>
                <m:cn>0</m:cn>
	          </m:apply>
	        </m:mtd>
	      </m:mtr>
	    </m:mtable>
	  </m:math>

	Note that the current entering a node is the negative of the
	current leaving the node.
      </para>
      
      <para id="para2wra">
	Given any two of these KCL equations, we can find the other by
	adding or subtracting them. Thus, one of them is redundant
	and, in mathematical terms, we can discard any one of
	them. The convention is to discard the equation for the
	(unlabeled) node at the bottom of the circuit.
      </para>


      <figure id="simplecircuitroar" orient="horizontal">
	<subfigure id="id5409075">
	  <media id="id2516092" alt=""><image src="circuit4.png" mime-type="image/png"/></media>
	</subfigure>
	<subfigure id="id8116141">
	  <media id="id8356115" alt=""><image src="circuit4a.png" mime-type="image/png"/></media>
	</subfigure> 
	<caption>
	  The circuit shown is perhaps the simplest circuit that
	  performs a signal processing function. The input is provided
	  by the voltage source
	  <m:math>
	    <m:ci>
	      <m:msub>
		<m:mi>v</m:mi>
		<m:mtext>in</m:mtext>
	      </m:msub>
	    </m:ci>
	  </m:math>	  
	  and the output is the voltage 
	  <m:math>
	    <m:ci>
	      <m:msub>
		<m:mi>v</m:mi>
		<m:mtext>out</m:mtext>
	      </m:msub>
	    </m:ci>
	  </m:math> 
	  across the resistor labelled
	  <m:math>
	    <m:ci>
	      <m:msub>
		<m:mi>R</m:mi>
		<m:mn>2</m:mn>
	      </m:msub>
	    </m:ci>
	  </m:math>.
	</caption>  
      </figure>

      <exercise id="exer1">
	<problem id="id7439851">
	  <para id="exer1a">
	    In writing KCL equations, you will find that in an
	    <m:math><m:ci>n</m:ci></m:math>-node circuit, exactly one
	    of them is always redundant. Can you sketch a proof of why
	    this might be true? Hint: It has to do with the fact that
	    charge won't accumulate in one place on its own.
	  </para>
	</problem>
	<solution id="id7259808">
	  <para id="exer1b">
	    KCL says that the sum of currents entering or leaving a
	    node must be zero. If we consider two nodes together as a
	    "supernode", KCL applies as well to currents entering the
	    combination.  Since no currents enter an entire circuit,
	    the sum of currents must be zero.  If we had a two-node
	    circuit, the KCL equation of one <emphasis>must</emphasis>
	    be the negative of the other, We can combine all but one
	    node in a circuit into a supernode; KCL for the supernode
	    must be the negative of the remaining node's KCL equation.
	    Consequently, specifying <m:math> <m:apply> <m:minus/>
	    <m:ci>n</m:ci> <m:cn>1</m:cn> </m:apply> </m:math> KCL
	    equations always specifies the remaining one.
	  </para>
	</solution>
      </exercise>
    </section>

    <section id="voltage">
      <title>Kirchoff's Voltage Law (KVL)</title>
      <para id="para3wra">
	The voltage law says that the sum of voltages around every
	closed loop in the circuit must equal zero. A closed loop has
	the obvious definition: Starting at a node, trace a path
	through the circuit that returns you to the origin node. KVL
	expresses the fact that electric fields are conservative: The
	total work performed in moving a test charge around a closed
	path is zero. The KVL equation for our circuit is

	<m:math display="block">
	    <m:apply>
	      <m:eq/> 
              <m:apply>
                <m:minus/>
		<m:apply>
		  <m:plus/>
		  <m:ci>
		    <m:msub>
		      <m:mi>v</m:mi>
		      <m:mn>1</m:mn>
		    </m:msub>
		  </m:ci>
		  <m:ci>
		    <m:msub>
		      <m:mi>v</m:mi>
		      <m:mn>2</m:mn>
		    </m:msub>
		  </m:ci>
		</m:apply>
		<m:ci>v</m:ci>
              </m:apply>
              <m:cn>0</m:cn>
	    </m:apply>
	  </m:math>

	In writing KVL equations, we follow the convention that an
	element's voltage enters with a plus sign when traversing the
	closed path, we go from the positive to the negative of the
	voltage's definition.
      </para>
    </section>

    <figure id="simplecircuitwra" orient="horizontal">
      <subfigure id="id6596737"> 
	<media id="id1165855116977" alt=""><image src="circuit4.png" mime-type="image/png"/></media>
      </subfigure>
      <subfigure id="id4100821"> 
	<media id="id1165856298293" alt=""><image src="circuit4a.png" mime-type="image/png"/></media>
      </subfigure> 
      <caption>
	The circuit shown is perhaps the simplest circuit that
	performs a signal processing function. The input is provided
	by the voltage source
	<m:math>
	  <m:ci>
	    <m:msub>
	      <m:mi>v</m:mi>
	      <m:mtext>in</m:mtext>
	    </m:msub>
	  </m:ci>
	</m:math> 
	and the output is the voltage
	<m:math>
	  <m:ci>
	    <m:msub>
	      <m:mi>v</m:mi>
	      <m:mtext>out</m:mtext>
	    </m:msub>
	  </m:ci>
	</m:math> 
	across the resistor labelled 
	<m:math>
	  <m:ci>
	    <m:msub>
	      <m:mi>R</m:mi>
	      <m:mn>2</m:mn>
	    </m:msub>
	  </m:ci>
	</m:math>. 
      </caption>
    </figure>

    <para id="para1roar">
      For the <link target-id="simplecircuitwra" strength="3">example
      circuit</link>, we have three
      <emphasis>v-i</emphasis> relations, two KCL equations, and
      one KVL equation for solving for the circuit's six voltages and
      currents.
     
 	<m:math display="block">
      <m:mtable>
        <m:mtr>
          <m:mtd>
            <m:mtext>v-i:</m:mtext>
          </m:mtd>
          <m:mtd>
	        <m:apply><m:eq/>
              <m:ci>v</m:ci>
              <m:ci><m:msub><m:mi>v</m:mi><m:mtext>in</m:mtext></m:msub></m:ci>
	        </m:apply>
	      </m:mtd>
	    </m:mtr>

	    <m:mtr>
	      <m:mtd/>
	      <m:mtd>
	        <m:apply><m:eq/>
              <m:ci><m:msub><m:mi>v</m:mi><m:mn>1</m:mn></m:msub></m:ci>
              <m:apply><m:times/>
	            <m:ci><m:msub><m:mi>R</m:mi><m:mn>1</m:mn></m:msub></m:ci>
	            <m:ci><m:msub><m:mi>i</m:mi><m:mn>1</m:mn></m:msub></m:ci>
              </m:apply>
	        </m:apply>
	      </m:mtd>
	    </m:mtr>

	    <m:mtr>
	      <m:mtd/>
	      <m:mtd>
	        <m:apply><m:eq/>
              <m:ci><m:msub><m:mi>v</m:mi><m:mtext>out</m:mtext></m:msub></m:ci>
              <m:apply><m:times/>
	            <m:ci><m:msub><m:mi>R</m:mi><m:mn>2</m:mn></m:msub></m:ci>
	            <m:ci><m:msub><m:mi>i</m:mi><m:mtext>out</m:mtext></m:msub></m:ci>
              </m:apply>
	        </m:apply>
	      </m:mtd>
	    </m:mtr>

        <m:mtr>
          <m:mtd><m:mtext>KCL:</m:mtext></m:mtd>
          <m:mtd>
	        <m:apply><m:eq/>
              <m:apply><m:minus/>
	            <m:apply><m:minus/><m:ci>i</m:ci></m:apply>
	            <m:ci><m:msub><m:mi>i</m:mi><m:mn>1</m:mn></m:msub></m:ci>
              </m:apply>
              <m:cn>0</m:cn>
	        </m:apply>
	      </m:mtd>
	    </m:mtr>

	    <m:mtr>
	      <m:mtd/>
	      <m:mtd>
	        <m:apply><m:eq/>
	          <m:apply><m:minus/>
	            <m:ci><m:msub><m:mi>i</m:mi><m:mn>1</m:mn></m:msub></m:ci>
	            <m:ci><m:msub><m:mi>i</m:mi><m:mtext>out</m:mtext></m:msub></m:ci>
	          </m:apply>
	          <m:cn>0</m:cn>
	        </m:apply>
	      </m:mtd>
	    </m:mtr>

        <m:mtr>
          <m:mtd><m:mtext>KVL:</m:mtext></m:mtd> 
          <m:mtd>
            <m:apply><m:eq/>
	          <m:apply><m:plus/>
	            <m:apply><m:minus/><m:ci>v</m:ci></m:apply>
	            <m:ci><m:msub><m:mi>v</m:mi><m:mn>1</m:mn></m:msub></m:ci>
	            <m:ci><m:msub><m:mi>v</m:mi><m:mtext>out</m:mtext></m:msub></m:ci>
	          </m:apply>
	          <m:cn>0</m:cn>
	        </m:apply>
	      </m:mtd>
	    </m:mtr>
      </m:mtable>
	</m:math>

    We have exactly the right number of equations! Eventually, we
      will discover shortcuts for solving circuit problems; for now,
      we want to eliminate all the variables but
      <m:math>
	<m:ci>
	  <m:msub>
	    <m:mi>v</m:mi>
	    <m:mtext>out</m:mtext>
	  </m:msub>
	</m:ci>
      </m:math> and determine how it depends on
      <m:math>
	<m:ci>
	  <m:msub>
	    <m:mi>v</m:mi>
	    <m:mtext>in</m:mtext>
	  </m:msub>
	</m:ci>
      </m:math> and on resistor values. 
      The KVL equation can be rewritten as  
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>
	    <m:msub>
	      <m:mi>v</m:mi>
	      <m:mtext>in</m:mtext>
	    </m:msub>
	  </m:ci>
	  <m:apply>
	    <m:plus/>
	    <m:ci>
	      <m:msub>
		<m:mi>v</m:mi>
		<m:mn>1</m:mn>
	      </m:msub>
	    </m:ci>
	    <m:ci>
	      <m:msub>
		<m:mi>v</m:mi>
		<m:mtext>out</m:mtext>
	      </m:msub>
	    </m:ci>
	  </m:apply>
	</m:apply>
      </m:math>.
      Substituting into it the resistor's <emphasis>v-i</emphasis>
      relation, we have
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>
	    <m:msub>
	      <m:mi>v</m:mi>
	      <m:mtext>in</m:mtext>
	    </m:msub>
	  </m:ci>
	  <m:apply>
	    <m:plus/>
	    <m:apply>
	      <m:times/>
	      <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mn>1</m:mn>
		</m:msub>
	      </m:ci>
	      <m:ci>
		<m:msub>
		  <m:mi>i</m:mi>
		  <m:mn>1</m:mn>
		</m:msub>
	      </m:ci>
	    </m:apply>
	    <m:apply>
	      <m:times/>
	      <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mn>2</m:mn>
		</m:msub>
	      </m:ci>
	      <m:ci>
		<m:msub>
		  <m:mi>i</m:mi>
		  <m:mtext>out</m:mtext>
		</m:msub>
	      </m:ci>
	    </m:apply>
          </m:apply>
	</m:apply>
      </m:math>.
      Yes, we temporarily eliminate the quantity we seek. Though not
      obvious, it is the simplest way to solve the equations. One of
      the KCL equations says
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>
	    <m:msub>
	      <m:mi>i</m:mi>
	      <m:mn>1</m:mn>
	    </m:msub>
	  </m:ci>
	  <m:ci>
	    <m:msub>
	      <m:mi>i</m:mi>
	      <m:mtext>out</m:mtext>
	    </m:msub>
	  </m:ci>
	</m:apply>
      </m:math>,
      which means that  
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>
	    <m:msub>
	      <m:mi>v</m:mi>
	      <m:mtext>in</m:mtext>
	    </m:msub>
	  </m:ci>
	  <m:apply>
	    <m:plus/>
	    <m:apply>
	      <m:times/>
	      <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mn>1</m:mn>
		</m:msub>
	      </m:ci>
	      <m:ci>
		<m:msub>
		  <m:mi>i</m:mi>
		  <m:mtext>out</m:mtext>
		</m:msub>
	      </m:ci>
	    </m:apply>
	    <m:apply>
	      <m:times/>
	      <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mn>2</m:mn>
		</m:msub>
	      </m:ci>
	      <m:ci>
		<m:msub>
		  <m:mi>i</m:mi>
		  <m:mtext>out</m:mtext>
		</m:msub>
	      </m:ci>
	    </m:apply>
	  </m:apply>
          <m:apply>
	    <m:times/>
	    <m:apply>
	      <m:plus/>
              <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mn>1</m:mn>
		</m:msub>
              </m:ci>
              <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mn>2</m:mn>
		</m:msub>
              </m:ci>
	    </m:apply>
	    <m:ci>
	      <m:msub>
		<m:mi>i</m:mi>
		<m:mtext>out</m:mtext>
	      </m:msub>
	    </m:ci>
          </m:apply>
	</m:apply>
      </m:math>.
      Solving for the current in the output resistor, we have  
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>
	    <m:msub>
	      <m:mi>i</m:mi>
	      <m:mtext>out</m:mtext>
	    </m:msub>
	  </m:ci>
	  <m:apply>
	    <m:divide/>
	    <m:ci>
	      <m:msub>
		<m:mi>v</m:mi>
		<m:mtext>in</m:mtext>
	      </m:msub>
	    </m:ci>
	    <m:apply>
	      <m:plus/>
	      <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mn>1</m:mn>
		</m:msub>
	      </m:ci>
	      <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mn>2</m:mn>
		</m:msub>
	      </m:ci>
	    </m:apply>
	  </m:apply>
	</m:apply>
      </m:math>.
      <emphasis>We have now solved the circuit</emphasis>: We have
      expressed one voltage or current in terms of sources and
      circuit-element values. To find any other circuit quantities, we
      can back substitute this answer into our original equations or
      ones we developed along the way. Using the
      <emphasis>v-i</emphasis> relation for the output resistor,
      we obtain the quantity we seek.

	<m:math display="block">
	  <m:apply>
	    <m:eq/>
	    <m:ci>
	      <m:msub>
		<m:mi>v</m:mi>
		<m:mtext>out</m:mtext>
	      </m:msub>
	    </m:ci>
	    <m:apply>
	      <m:times/>
	      <m:apply>
		<m:divide/>
		<m:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mn>2</m:mn>
		  </m:msub>
		</m:ci>
		<m:apply>
		  <m:plus/>
		  <m:ci>
		    <m:msub>
		      <m:mi>R</m:mi>
		      <m:mn>1</m:mn>
		    </m:msub>
		  </m:ci>
		  <m:ci>
		    <m:msub>
		      <m:mi>R</m:mi>
		      <m:mn>2</m:mn>
		    </m:msub>
		  </m:ci>
		</m:apply>
	      </m:apply>
	      <m:ci>
		<m:msub>
		  <m:mi>v</m:mi>
		  <m:mtext>in</m:mtext>
		</m:msub>
	      </m:ci>
	    </m:apply>
	  </m:apply>
	</m:math>
    </para>

    <exercise id="exer1gwar">
      <problem id="id2269821"> 
	<para id="exer1agwar">
	  Referring back to 
	  <link target-id="simplecircuit" strength="3"/>, 
	  a circuit should serve some useful purpose. What kind of
	  system does our circuit realize and, in terms of element
	  values, what are the system's parameter(s)?
	</para>
      </problem>

      <solution id="id5481042">
	<para id="exer1bgwar">
	  The circuit serves as an amplifier having a gain of  
	  <m:math>
	    <m:apply>
	      <m:divide/>
	      <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mn>2</m:mn>
		</m:msub>
	      </m:ci>
	      <m:apply>
		<m:plus/>
		<m:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mn>1</m:mn>
		  </m:msub>
		</m:ci>
		<m:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mn>2</m:mn>
		  </m:msub>
		</m:ci>
	      </m:apply>
	    </m:apply>
	  </m:math>.
	</para>
      </solution>
    </exercise>

  </content>
</document>
