<?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="m0020" module-id="" cnxml-version="0.6"> 

  <title>Equivalent Circuits: Resistors and Sources</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>m0020</md:content-id>
  <md:title>Equivalent Circuits: Resistors and Sources</md:title>
  <md:version>2.23</md:version>
  <md:created>2000/07/07</md:created>
  <md:revised>2009/06/04 11:35:38.674 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="liqun">
        <md:firstname>Liqun</md:firstname>
        <md:surname>Wang</md:surname>
        <md:fullname>Liqun Wang</md:fullname>
        <md:email>liqun@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>Equivalent Circuits</md:keyword>
    <md:keyword>Mayer-Norton</md:keyword>
    <md:keyword>Norton</md:keyword>
    <md:keyword>Thevenin</md:keyword>
  </md:keywordlist>
  <md:subjectlist>
    <md:subject>Science and Technology</md:subject>
  </md:subjectlist>
  <md:abstract>Introduction of equivalent circuits, including the function in v-i relation.</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">
      We have found that the way to think about circuits is to locate
      and group parallel and series resistor combinations. Those
      resistors not involved with variables of interest can be
      collapsed into a single resistance. This result is known as an
      <term>equivalent circuit</term>: from the viewpoint of a pair of
      terminals, a group of resistors functions as a single resistor,
      the resistance of which can usually be found by applying the
      parallel and series rules.
    </para>
    
    <figure id="eqcircuit">
      
      <media id="id15967562" alt="">
        <image src="circuit12.png" mime-type="image/png"/>
        <image src="circuit12.eps" mime-type="application/postscript"/>
      </media>
    </figure>

    <para id="para2">
      This result generalizes to include sources in a very interesting
      and useful way. Let's consider our simple attenuator circuit (shown in the <link target-id="eqcircuit">figure</link>)
      from the viewpoint of the output terminals. We want to find the
      <emphasis>v-i</emphasis> relation for the output terminal pair,
      and then find the equivalent circuit for the boxed circuit. To
      perform this calculation, use the circuit laws and element
      relations, but do not attach anything to the output
      terminals. We seek the relation between
      <m:math><m:ci>v</m:ci></m:math> and
      <m:math><m:ci>i</m:ci></m:math> that describes the kind of
      element that lurks within the dashed box. The result is
      <equation id="eq1">
	<m:math>
	  <m:apply>
	    <m:eq/>
	    <m:ci>v</m:ci>
	    <m:apply>
	      <m:plus/>
	      <m:apply>
		<m:times/>
		<m:apply>
		  <m:ci><m:mo>∥</m:mo></m:ci>
		  <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>i</m:ci>
	      </m:apply>
	      <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:mi>in</m:mi>
		  </m:msub>
		</m:ci>
	      </m:apply>
	    </m:apply>
	  </m:apply>
	</m:math>
      </equation>
      If the source were zero, it could be replaced by a short
      circuit, which would confirm that the circuit does indeed
      function as a parallel combination of resistors.  However, the
      source's presence means that the circuit is
      <emphasis>not</emphasis> well modeled as a resistor.  
    </para>
    
    <figure id="thevenin">
      
      <media id="id3729529" alt="">
        <image src="thevenin.png" mime-type="image/png"/>
        <image src="thevenin.eps" mime-type="application/postscript"/>
      </media>
      <caption>
	The Thévenin equivalent circuit.
      </caption>
    </figure> 

    <para id="para3">
      If we consider the simple circuit of <link target-id="thevenin" strength="2"/>, we find it has the <emphasis>v-i</emphasis>
	relation at its terminals of
      <equation id="eq2">
	<m:math>
	  <m:apply>
	    <m:eq/>
	    <m:ci>v</m:ci>
	    <m:apply>
	      <m:plus/>
	      <m:apply>
		<m:times/>
		<m:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mi>eq</m:mi>
		  </m:msub>
		</m:ci>
		<m:ci>i</m:ci>
	      </m:apply>
	      <m:ci>
		<m:msub>
		  <m:mi>v</m:mi>
		  <m:mi>eq</m:mi>
		</m:msub>
	      </m:ci>
	    </m:apply>
	  </m:apply>
	</m:math>
      </equation> 

      Comparing the two <emphasis>v-i</emphasis> relations, we find
      that they have the same form. In this case the
      <emphasis>Thévenin equivalent resistance</emphasis> is
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>
	    <m:msub>
	      <m:mi>R</m:mi>
	      <m:mi>eq</m:mi>
	    </m:msub>
	  </m:ci>
	  <m:apply>
	    <m:mo>∥</m:mo>
	    <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>      
      and the  <emphasis>Thévenin equivalent source</emphasis>
      has voltage  
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>
	    <m:msub>
	      <m:mi>v</m:mi>
	      <m:mi>eq</m:mi>
	    </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:mi>in</m:mi>
	      </m:msub>
	    </m:ci>
	  </m:apply>
	</m:apply>
      </m:math>.  
      Thus, from viewpoint of the terminals, you cannot distinguish the
      two circuits. Because the equivalent circuit has fewer elements,
      it is easier to analyze and understand than any other
      alternative.  
    </para>

    <para id="para2rub"> 
      For <emphasis>any</emphasis> circuit containing resistors and
      sources, the <emphasis>v-i</emphasis> relation will be of the
      form

      <equation id="eq1rub">
	<m:math>
	  <m:apply>
	    <m:eq/>
	    <m:ci>v</m:ci>
	    <m:apply>
	      <m:plus/>
	      <m:apply>
		<m:times/>
		<m:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mi>eq</m:mi>
		  </m:msub>
		</m:ci>
		<m:ci>i</m:ci>
	      </m:apply>
	      <m:ci>
		<m:msub>
		  <m:mi>v</m:mi>
		  <m:mi>eq</m:mi>
		</m:msub>
	      </m:ci>
	    </m:apply>
	  </m:apply>
	</m:math>
      </equation>

      and the <term>Thévenin equivalent circuit</term> for any
      such circuit is that of <link target-id="thevenin" strength="2"/>. This equivalence applies no
      matter how many sources or resistors may be present in the
      circuit. In the <link target-id="ex1" strength="2">example</link> below,
      we know the circuit's construction and element values, and
      derive the equivalent source and resistance.  Because
      Thévenin's theorem applies in general, we should be able
      to make measurements or calculations <emphasis>only from the
	terminals</emphasis> to determine the equivalent circuit.
    </para>


    <para id="para3rub">
      To be more specific, consider the equivalent circuit of <link target-id="thevenin" strength="2">this figure</link>.  Let the
      terminals be open-circuited, which has the effect of setting the
      current <m:math><m:ci>i</m:ci></m:math> to zero.  Because no
      current flows through the resistor, the voltage across it is
      zero (remember, Ohm's Law says that
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>v</m:ci>
	  <m:apply>
	    <m:times/>
	    <m:ci>R</m:ci>
	    <m:ci>i</m:ci>
	  </m:apply>
	</m:apply>
      </m:math>).  
      Consequently, by applying KVL we have that the so-called
      open-circuit voltage
      <m:math>
	<m:ci>
	  <m:msub>
	    <m:mi>v</m:mi>
	    <m:mi>oc</m:mi>
	  </m:msub>
	</m:ci>
      </m:math>
      equals the Thévenin equivalent voltage. Now consider the
      situation when we set the terminal voltage to zero
      (short-circuit it) and measure the resulting current. Referring
      to the equivalent circuit, the source voltage now appears
      entirely across the resistor, leaving the short-circuit current
      to be
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>
	    <m:msub>
	      <m:mi>i</m:mi>
	      <m:mi>sc</m:mi>
	    </m:msub>
	  </m:ci>
	  <m:apply>
	    <m:minus/>
	    <m:apply>
	      <m:divide/>
	      <m:ci>
		<m:msub>
		  <m:mi>v</m:mi>
		  <m:mi>eq</m:mi>
		</m:msub>
	      </m:ci>
	      <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mi>eq</m:mi>
		</m:msub>
	      </m:ci>
	    </m:apply>
	  </m:apply>
	</m:apply>
      </m:math>.                                               
      From this property, we can determine the equivalent resistance.
    </para>   

    <equation id="eq2rub">
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>
	    <m:msub>
	      <m:mi>v</m:mi>
	      <m:mi>eq</m:mi>
	    </m:msub>
	  </m:ci>
	  <m:ci>
	    <m:msub>
	      <m:mi>v</m:mi>
	      <m:mi>oc</m:mi>
	    </m:msub>
	  </m:ci>
	</m:apply>
      </m:math>
    </equation>

    <equation id="eq3">
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>
	    <m:msub>
	      <m:mi>R</m:mi>
	      <m:mi>eq</m:mi>
	    </m:msub>
	  </m:ci>
	  <m:apply>
	    <m:minus/>
	    <m:apply>
	      <m:divide/>
	      <m:ci>
		<m:msub>
		  <m:mi>v</m:mi>
		  <m:mi>oc</m:mi>
		</m:msub>
	      </m:ci>
	      <m:ci>
		<m:msub>
		  <m:mi>i</m:mi>
		  <m:mi>sc</m:mi>
		</m:msub>
	      </m:ci>
	    </m:apply>
	  </m:apply>
	</m:apply>
      </m:math>
    </equation>

    <exercise id="exer1">
      <problem id="id4221673">
	<para id="exer1a">
	  Use the open/short-circuit approach to derive the
	  Thévenin equivalent of the circuit shown in <link target-id="tworesist" strength="2"/>.
	</para>

	<figure id="tworesist">
	  <media id="id3956347" alt=""><image src="circuit12.png" mime-type="image/png"/></media>
	</figure>
      </problem>

      <solution id="id4320251">
	<para id="exer1b">
	  <m:math>
	    <m:apply>
	      <m:eq/>
	      <m:ci>
		<m:msub>
		  <m:mi>v</m:mi>
		  <m:mi>oc</m:mi>
		</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:mi>in</m:mi>
		  </m:msub>
		</m:ci>
	      </m:apply>
	    </m:apply>
	  </m:math> and
	  
	  <m:math>
	    <m:apply>
	      <m:eq/>
	      <m:ci>
		<m:msub>
		  <m:mi>i</m:mi>
		  <m:mi>sc</m:mi>
		</m:msub>
	      </m:ci>
	      <m:apply>
		<m:minus/>
		<m:apply>
		  <m:divide/>
		  <m:ci>
		    <m:msub>
		      <m:mi>v</m:mi>
		      <m:mi>in</m:mi>
		    </m:msub>
		  </m:ci>
		  <m:ci>
		    <m:msub>
		      <m:mi>R</m:mi>
		      <m:mn>1</m:mn>
		    </m:msub>
		  </m:ci>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:math>
	  (resistor
	  <m:math>
	    <m:ci>
	      <m:msub>
		<m:mi>R</m:mi>
		<m:mn>2</m:mn>
	      </m:msub>
	    </m:ci>
	  </m:math> 	  
	  is shorted out in this case). Thus,
	  <m:math>
	    <m:apply>
	      <m:eq/>
	      <m:ci>
		<m:msub>
		  <m:mi>v</m:mi>
		  <m:mi>eq</m:mi>
		</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:mi>in</m:mi>
		  </m:msub>
		</m:ci>
	      </m:apply>
	    </m:apply>
	  </m:math> 
	  and 
	  <m:math>
	    <m:apply>
	      <m:eq/>
	      <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mi>eq</m:mi>
		</m:msub>
	      </m:ci>
	      <m:apply>
		<m:divide/>
		<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>R</m:mi>
		      <m:mn>2</m:mn>
		    </m:msub>
		  </m:ci>
		</m:apply>
		<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>.
	</para>
      </solution>
    </exercise>
    

    <example id="ex1"> 
      <figure id="Thevenincircuit">
	<media id="id4007487" alt=""><image src="circuit13.png" mime-type="image/png"/></media>
      </figure> 

      <para id="para4">For the circuit depicted in <link target-id="Thevenincircuit" strength="2"/>, let's derive its Thévenin
	equivalent two different ways. Starting with the
	open/short-circuit approach, let's first find the open-circuit
	voltage
	<m:math>
	  <m:ci>
	    <m:msub>
	      <m:mi>v</m:mi>
	      <m:mi>oc</m:mi>
	    </m:msub>
	  </m:ci>
	</m:math>. 
	We have a current divider relationship as
	<m:math>
	  <m:ci>
	    <m:msub>
	      <m:mi>R</m:mi>
	      <m:mn>1</m:mn>
	    </m:msub>
	  </m:ci>
	</m:math>
	is in parallel with the series combination of
	<m:math>
	  <m:ci>
	    <m:msub>
	      <m:mi>R</m:mi>
	      <m:mn>2</m:mn>
	    </m:msub>
	  </m:ci>
	</m:math> and
	<m:math>
	  <m:ci>
	    <m:msub>
	      <m:mi>R</m:mi>
	      <m:mn>3</m:mn>
	    </m:msub>
	  </m:ci>
	</m:math>.
	Thus,  
	<m:math>
	  <m:apply>
	    <m:eq/>
	    <m:ci>
	      <m:msub>
		<m:mi>v</m:mi>
		<m:mi>oc</m:mi>
	      </m:msub>
	    </m:ci>
	    <m:apply>
	      <m:divide/>
	      <m:apply>
		<m:times/>
		<m:ci>
		  <m:msub>
		    <m:mi>i</m:mi>
		    <m:mi>in</m:mi>
		  </m:msub>
		</m:ci>
		<m:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mn>3</m:mn>
		  </m:msub>
		</m:ci>
		<m:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mn>1</m:mn>
		  </m:msub>
		</m:ci>
	      </m:apply>
	      <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:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mn>3</m:mn>
		  </m:msub>
		</m:ci>
	      </m:apply>
	    </m:apply>
	  </m:apply>
	</m:math>.
	When we short-circuit the terminals, no voltage appears across 
	<m:math>
	  <m:ci>
	    <m:msub>
	      <m:mi>R</m:mi>
	      <m:mn>3</m:mn>
	    </m:msub>
	  </m:ci>
	</m:math>, 	
	and thus no current flows through it. In short,
	<m:math>
	  <m:ci>
	    <m:msub>
	      <m:mi>R</m:mi>
	      <m:mn>3</m:mn>
	    </m:msub>
	  </m:ci>
	</m:math>
	does not affect the short-circuit current, and can be eliminated. We
	again have a current divider relationship:  
	<m:math>
	  <m:apply>
	    <m:eq/>
	    <m:ci>
	      <m:msub>
		<m:mi>i</m:mi>
		<m:mi>sc</m:mi>
	      </m:msub>
	    </m:ci>
	    <m:apply>
	      <m:minus/>
	      <m:apply>
		<m:divide/>
		<m:apply>
		  <m:times/>
		  <m:ci>
		    <m:msub>
		      <m:mi>i</m:mi>
		      <m:mi>in</m:mi>
		    </m:msub>
		  </m:ci>
		  <m:ci>
		    <m:msub>
		      <m:mi>R</m:mi>
		      <m:mn>1</m:mn>
		    </m:msub>
		  </m:ci>
		</m:apply>
		<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:apply>
	</m:math>. 
	Thus, the Thévenin equivalent resistance is
	<m:math>
	  <m:apply>
	    <m:divide/>
	    <m:apply>
	      <m:times/>
	      <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mn>3</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: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:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mn>3</m:mn>
		</m:msub>
	      </m:ci>
	    </m:apply>
	  </m:apply>
	</m:math>.
      </para>

      <para id="para5">
	To verify, let's find the equivalent resistance by reaching
	inside the circuit and setting the current source to
	zero. Because the current is now zero, we can replace the
	current source by an open circuit. From the viewpoint of the
	terminals, resistor
	<m:math>
	  <m:ci>
	    <m:msub>
	      <m:mi>R</m:mi>
	      <m:mn>3</m:mn>
	    </m:msub>
	  </m:ci>
	</m:math> 
	is now in parallel with the series combination of
	<m:math>
	  <m:ci>
	    <m:msub>
	      <m:mi>R</m:mi>
	      <m:mn>1</m:mn>
	    </m:msub>
	  </m:ci>
	</m:math> and
	<m:math>
	  <m:ci>
	    <m:msub>
	      <m:mi>R</m:mi>
	      <m:mn>2</m:mn>
	    </m:msub>
	  </m:ci>
	</m:math>.
	Thus,   
	<m:math>
	  <m:apply>
	    <m:eq/>
	    <m:ci>
	      <m:msub>
		<m:mi>R</m:mi>
		<m:mi>eq</m:mi>
	      </m:msub>
	    </m:ci>
	    <m:apply>
	      <m:ci><m:mo>∥</m:mo></m:ci>
	      <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mn>3</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:apply>
	</m:math>,
	and we obtain the same result.
      </para>
    </example>


    <figure id="equivalents">
      <media id="id3487834" alt="">
        <image src="equivalents.png" mime-type="image/png"/>
        <image src="equivalents.eps" mime-type="application/postscript"/>
      </media>
      <caption>
	<emphasis>All</emphasis> circuits containing sources and
	resistors can be described by simpler equivalent
	circuits. Choosing the one to use depends on the application,
	not on what is actually inside the circuit.
      </caption>
    </figure> 

    <para id="para1zap">
      As you might expect, equivalent circuits come in two forms: the
      voltage-source oriented <link document="m0021" strength="2">Thévenin equivalent</link> and the
      current-source oriented <term>Mayer-Norton equivalent</term>
      (<link target-id="equivalents" strength="2"/>).
      To derive the latter, the  <emphasis>v-i</emphasis> relation for
      the Thévenin equivalent can be written as
      
      <equation id="eq1zap">
	<m:math>
	  <m:apply>
	    <m:eq/>
	    <m:ci>v</m:ci>
	    <m:apply>
	      <m:plus/>
	      <m:apply>
		<m:times/>
		<m:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mi>eq</m:mi>
		  </m:msub>
		</m:ci>
		<m:ci>i</m:ci>
	      </m:apply>
	      <m:ci>
		<m:msub>
		  <m:mi>v</m:mi>
		  <m:mi>eq</m:mi>
		</m:msub>
	      </m:ci>
	    </m:apply>
	  </m:apply>
	</m:math> 
      </equation>
      or
      <equation id="eq2zap">
	<m:math>
	  <m:apply>
	    <m:eq/>
	    <m:ci>i</m:ci>
	    <m:apply>
	      <m:minus/>
	      <m:apply>
		<m:divide/>
		<m:ci>v</m:ci>
		<m:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mi>eq</m:mi>
		  </m:msub>
		</m:ci>
	      </m:apply>
	      <m:ci>
		<m:msub>
		  <m:mi>i</m:mi>
		  <m:mi>eq</m:mi>
		</m:msub>
	      </m:ci>
	    </m:apply>
	  </m:apply>
	</m:math>
      </equation>
      where
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>
	    <m:msub>
	      <m:mi>i</m:mi>
	      <m:mi>eq</m:mi>
	    </m:msub>
	  </m:ci>
	  <m:apply>
	    <m:divide/>
	    <m:ci>
	      <m:msub>
		<m:mi>v</m:mi>
		<m:mi>eq</m:mi>
	      </m:msub>
	    </m:ci>
	    <m:ci>
	      <m:msub>
		<m:mi>R</m:mi>
		<m:mi>eq</m:mi>
	      </m:msub>
	    </m:ci>
	  </m:apply>
	</m:apply>
      </m:math>
      is the Mayer-Norton equivalent source. The Mayer-Norton
      equivalent shown in <link target-id="equivalents" strength="2"/> be
      easily shown to have this <emphasis>v-i</emphasis>
      relation. Note that both variations have the same equivalent
      resistance. The short-circuit current equals the negative of the
      Mayer-Norton equivalent source.
    </para>


    <exercise id="exer1zap">
      <problem id="id13975188">
	<para id="exer1azap">
	  Find the Mayer-Norton equivalent circuit for the circuit below.
	</para>

	<figure id="Thevenincircuitrub">
	  <media id="id3135554" alt="">
            <image src="circuit13.png" mime-type="image/png"/>
            <image src="circuit13.eps" mime-type="application/postscript"/>
          </media>
	</figure> 
      </problem>

      <solution id="id11756024">
	<para id="exer1bzap">
	  <m:math>
	    <m:apply>
	      <m:eq/>
	      <m:ci>
		<m:msub>
		  <m:mi>i</m:mi>
		  <m:mi>eq</m:mi>
		</m:msub>
	      </m:ci>
	      <m:apply>
		<m:times/>
		<m:apply>
		  <m:divide/>
		  <m:ci>
		    <m:msub>
		      <m:mi>R</m:mi>
		      <m:mn>1</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>i</m:mi>
		    <m:mi>in</m:mi>
		  </m:msub>
		</m:ci>
	      </m:apply>
	    </m:apply>
	  </m:math>
	  and  
	  <m:math>
	    <m:apply>
	      <m:eq/>
	      <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mi>eq</m:mi>
		</m:msub>
	      </m:ci>
	      <m:apply>
		<m:ci><m:mo>∥</m:mo></m:ci>
		<m:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mn>3</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:apply>
	  </m:math>.
	</para>
      </solution>
    </exercise>


    <para id="para3stun">       
      Equivalent circuits can be used in two basic ways. The first is
      to simplify the analysis of a complicated circuit by realizing
      the <emphasis>any</emphasis> portion of a circuit can be
      described by either a Thévenin or Mayer-Norton
      equivalent. Which one is used depends on whether what is
      attached to the terminals is a series configuration (making the
      Thévenin equivalent the best) or a parallel one (making
      Mayer-Norton the best).
    </para>
    
    <para id="para4stun"> 
      Another application is modeling. When we buy a flashlight
      battery, either equivalent circuit can accurately describe
      it. These models help us understand the limitations of a
      battery. Since batteries are labeled with a voltage
      specification, they should serve as voltage sources and the
      Thévenin equivalent serves as the natural choice. If a
      load resistance
      <m:math>
	<m:ci>
	  <m:msub>
	    <m:mi>R</m:mi>
	    <m:mi>L</m:mi>
	  </m:msub>
	</m:ci>
      </m:math> 
      is placed across its terminals, the voltage output can be found using
      voltage divider:  
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>v</m:ci>
	  <m:apply>
	    <m:divide/>
	    <m:apply>
	      <m:times/>
	      <m:ci>
		<m:msub>
		  <m:mi>v</m:mi>
		  <m:mi>eq</m:mi>
		</m:msub>
	      </m:ci>
	      <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mi>L</m:mi>
		</m:msub>
	      </m:ci>
	    </m:apply>
	    <m:apply>
	      <m:plus/>
	      <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mi>L</m:mi>
		</m:msub>
	      </m:ci>
	      <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mi>eq</m:mi>
		</m:msub>
	      </m:ci>
	    </m:apply>
	  </m:apply>
	</m:apply>
      </m:math>. If we have a load resistance much larger than the
      battery's equivalent resistance, then, to a good approximation,
      the battery does serve as a voltage source. If the load
      resistance is much smaller, we certainly don't have a voltage
      source (the output voltage depends directly on the load
      resistance). Consider now the Mayer-Norton equivalent; the
      current through the load resistance is given by current divider,
      and equals
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>i</m:ci>
	  <m:apply>
	    <m:minus/>
	    <m:apply>
	      <m:divide/>
	      <m:apply>
		<m:times/>
		<m:ci>
		  <m:msub>
		    <m:mi>i</m:mi>
		    <m:mi>eq</m:mi>
		  </m:msub>
		</m:ci>
		<m:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mi>eq</m:mi>
		  </m:msub>
		</m:ci>
	      </m:apply>
	      <m:apply>
		<m:plus/>
		<m:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mi>L</m:mi>
		  </m:msub>
		</m:ci>
		<m:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mi>eq</m:mi>
		  </m:msub>
		</m:ci>
	      </m:apply>
	    </m:apply>
	  </m:apply>
	</m:apply>
      </m:math>.  For a current that does not vary with the load
      resistance, this resistance should be much smaller than the
      equivalent resistance. If the load resistance is comparable to
      the equivalent resistance, the battery serves
      <emphasis>neither</emphasis> as a voltage source or a current
      course. Thus, when you buy a battery, you get a voltage source
      if its equivalent resistance is much
      <emphasis>smaller</emphasis> than the equivalent resistance of
      the circuit to which you attach it. On the other hand, if you
      attach it to a circuit having a small equivalent resistance, you
      bought a current source.
    </para>

    <para id="history">
      <note id="id14424528" type="Léon Charles Thévenin"><label>Léon Charles Thévenin</label>
	He was an engineer with France's Postes,
	Télégraphe et Téléphone. In 1883,
	he published (twice!)  a proof of what is now called the
	Thévenin equivalent while developing ways of teaching
	electrical engineering concepts at the École
	Polytechnique.  He did not realize that the same result had
	been published by <link url="http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Helmholtz.html">Hermann
	Helmholtz</link>, the renowned nineteenth century physicist,
	thiry years earlier.
      </note>

      <note id="id5382598" type="Hans Ferdinand Mayer"><label>Hans Ferdinand Mayer</label>
	After earning his doctorate in physics in 1920, he turned to
	communications engineering when he joined Siemens &amp; Halske
	in 1922.  In 1926, he published in a German technical journal
	the Mayer-Norton equivalent. During his interesting career, he
	rose to lead Siemen's Central Laboratory in 1936,
	surruptiously leaked to the British all he knew of German
	warfare capabilities a month after the Nazis invaded Poland,
	was arrested by the Gestapo in 1943 for listening to BBC radio
	broadcasts, spent two years in Nazi concentration camps, and
	went to the United States for four years working for the Air
	Force and Cornell University before returning to Siemens in
	1950. He rose to a position on Siemen's Board of
	Directors before retiring.
      </note>

      <note id="id8370419" type="Edward L. Norton"><label>Edward L. Norton</label>
	<link url="http://www.ece.rice.edu/~dhj/norton">Edward
	Norton</link> was an electrical engineer who worked at Bell
	Laboratory from its inception in 1922.  In the
	<emphasis>same</emphasis> month when Mayer's paper appeared,
	Norton wrote in an internal technical memorandum a paragraph
	describing the current-source equivalent.  No evidence
	suggests Norton knew of Mayer's publication.
      </note>
    </para>
  </content>
</document>
