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<!DOCTYPE document PUBLIC "-//CNX//DTD CNXML 0.5 plus MathML//EN" "http://cnx.rice.edu/cnxml/0.5/DTD/cnxml_mathml.dtd">
<document xmlns="http://cnx.rice.edu/cnxml" xmlns:md="http://cnx.rice.edu/mdml/0.4" xmlns:m="http://www.w3.org/1998/Math/MathML" xmlns:bib="http://bibtexml.sf.net/" id="m10674">

  <name>Series and Parallel Circuits</name>

  <metadata>
  <md:version>2.7</md:version>
  <md:created>2002/06/19</md:created>
  <md:revised>2007/05/10 08:59:58.752 GMT-5</md:revised>
  <md:authorlist>
      <md:author id="dhj">
      <md:firstname>Don</md:firstname>
      
      <md:surname>Johnson</md:surname>
      <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:email>dhj@rice.edu</md:email>
    </md:maintainer>
    <md:maintainer id="bfite">
      <md:firstname>Benjamin</md:firstname>
      
      <md:surname>Fite</md:surname>
      <md:email>bfite@rice.edu</md:email>
    </md:maintainer>
  </md:maintainerlist>
  
  <md:keywordlist>
    <md:keyword>current divider</md:keyword>
    <md:keyword>input-output relationship</md:keyword>
    <md:keyword>load</md:keyword>
    <md:keyword>parallel</md:keyword>
    <md:keyword>series</md:keyword>
    <md:keyword>voltage divider</md:keyword>
  </md:keywordlist>

  <md:abstract>A breif look at series and parallel circuits.  Also defines
      voltage and current divider.</md:abstract>
</metadata>

  <content>
    <figure id="simplecircuitgwar" orient="horizontal">
      <subfigure>
	<media type="image/png" src="circuit4.png"/>
      </subfigure>
      <subfigure>
	<media type="image/png" src="circuit4a.png"/>
      </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:mi>in</m:mi>
	    </m:msub>
	  </m:ci>
	</m:math>
	and the output is the voltage
	<m:math>
	  <m:ci>
	    <m:msub>
	      <m:mi>v</m:mi>
	      <m:mi>out</m:mi>
	    </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="para1gwar">
      The results shown in other modules 
      (<cnxn document="m0014" strength="9">circuit elements</cnxn>, 
      <cnxn document="m0014" strength="9">KVL and KCL</cnxn>, 
      <cnxn document="m0014" strength="9">interconnection laws</cnxn>) 
      with regard to <cnxn target="simplecircuitgwar" strength="9">this
      circuit</cnxn>, and the values of other currents and voltages in this
      circuit as well, have profound implications.
      </para>

    <para id="voltdivide">
      Resistors connected in such a way that current from one must
      flow <emphasis>only</emphasis> into another—currents in
      all resistors connected this way have the same
      magnitude—are said to be connected in
      <term>series</term>.  For the two series-connected resistors in
      the example, <emphasis>the voltage across one resistor equals
      the ratio of that resistor's value and the sum of resistances
      times the voltage across the series combination</emphasis>. This
      concept is so pervasive it has a name: <term>voltage
      divider</term>.
    </para>
    
    <para id="iorelationship"> 
      The <term>input-output relationship</term> for this system,
      found in this particular case by voltage divider, takes the form
      of a ratio of the output voltage to the input voltage.

	<m:math display="block">
	  <m:apply>
	    <m:eq/>
            <m:apply>
	      <m:divide/>
	      <m:ci>
		<m:msub>
		  <m:mi>v</m:mi>
		  <m:mi>out</m:mi>
		</m:msub>
	      </m:ci>
	      <m:ci>
		<m:msub>
		  <m:mi>v</m:mi>
		  <m:mi>in</m:mi>
		</m:msub>
	      </m:ci>
	    </m:apply>
	    <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:apply>
	</m:math>

      In this way, we express how the components used to build the
      system affect the input-output relationship.  Because this
      analysis was made with ideal circuit elements, we might expect
      this relation to break down if the input amplitude is too high
      (Will the circuit survive if the input changes from 1 volt to
      one million volts?) or if the source's frequency becomes too
      high. In any case, this important way of expressing input-output
      relationships—as a ratio of output to
      input—pervades circuit and system theory.
    </para>
    
    <para id="currentwra">
      The current 
      <m:math>
        <m:ci>
          <m:msub>
            <m:mi>i</m:mi>
            <m:mn>1</m:mn>
          </m:msub>
        </m:ci>
      </m:math>
      is the current flowing out of the voltage source. Because it equals 
      <m:math>
	<m:ci>
	  <m:msub>
	    <m:mi>i</m:mi>
	    <m:mn>2</m:mn>
	  </m:msub>
	</m:ci>
      </m:math>, 
      we have that 
      <m:math>
	<m:apply>
	  <m:eq/>
          <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>i</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:math>:

      <note type="Resistors in series">
	The series combination of two resistors acts, as far as the
	voltage source is concerned, as a single resistor having a
	value equal to the sum of the two resistances.
      </note> 

      This result is the first of several <cnxn document="0020" strength="5">equivalent circuit ideas</cnxn>: In many cases, a
      complicated circuit when viewed from its terminals (the two
      places to which you might attach a source) appears to be a
      single circuit element (at best) or a simple combination of
      elements at worst. Thus, the equivalent circuit for a series
      combination of resistors is a single resistor having a
      resistance equal to the sum of its component resistances.
    </para>

    <figure id="eqcircuit">
      <media type="image/png" src="circuit4b.png"/>
      <caption>
	The resistor (on the right) is equivalent to the two resistors
	(on the left) and has a resistance equal to the sum of the
	resistances of the other two resistors.
      </caption>
    </figure>
    
    <para id="conclusion">
      Thus, the circuit the voltage source "feels" (through the
      current drawn from it) is a single resistor having resistance
      <m:math>
	<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:math>.
      Note that in making this equivalent circuit, the output voltage
      can no longer be defined: The output resistor labeled
      <m:math>
	<m:ci>
	  <m:msub>
	    <m:mi>R</m:mi>
	    <m:mn>2</m:mn>
	  </m:msub>
	</m:ci>
      </m:math>
      no longer appears. Thus, this equivalence is made strictly from
      the voltage source's viewpoint.
    </para>     

    <figure id="simpleparallel"> 
      <media type="image/png" src="circuit25.png"/>
      <caption>
	A simple parallel circuit.
      </caption>
    </figure>


     <para id="para1lit2">
      One interesting simple circuit 
      (<cnxn target="simpleparallel" strength="5"/>) 
      has two resistors connected side-by-side, what we will term a
      <term>parallel</term> connection, rather than in series.
      Here, applying KVL reveals that all the voltages are identical:
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>
	    <m:msub>
	      <m:mi>v</m:mi>
	      <m:mn>1</m:mn>
	    </m:msub>
	  </m:ci>
	  <m:ci>v</m:ci>
	</m:apply>
      </m:math>
      and
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>
	    <m:msub>
	      <m:mi>v</m:mi>
	      <m:mn>2</m:mn>
	    </m:msub>
	  </m:ci>
	  <m:ci>v</m:ci>
	</m:apply>
      </m:math>.
      This result typifies parallel connections. To write the KCL
      equation, note that the top node consists of the entire upper
      interconnection section. The KCL equation is
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:apply>
	    <m:minus/>
	    <m:apply>
	      <m:minus/>
	      <m:ci>
		<m:msub>
		  <m:mi>i</m:mi>
		  <m:mi>in</m:mi>
		</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: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:math>.
      Using the <emphasis>v-i</emphasis> relations, we find that
 
	<m:math display="block">
	  <m:apply>
	    <m:eq/>
	    <m:ci>
	      <m:msub>
		<m:mi>i</m:mi>
		<m:mi>out</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>
    </para>


    <exercise id="exer1wra">
      <problem>
	<para id="exer1awar">
	  Suppose that you replaced the current source in 
	  <cnxn target="simpleparallel" strength="5"/> 
	  by a voltage source.  How would 
	  <m:math>
	    <m:ci>
	      <m:msub>
		<m:mi>i</m:mi>
		<m:mi>out</m:mi>
	      </m:msub>
	    </m:ci>
	  </m:math>
	  be related to the source voltage?  Based on this result,
	  what purpose does this revised circuit have?
	</para>
      </problem>

      <solution>
	<para id="exer1bwra">
	  Replacing the current source by a voltage source does not
	  change the fact that the voltages are
	  identical. Consequently,
	  <m:math>
	    <m:apply>
	      <m:eq/>
              <m:ci>
                <m:msub>
                  <m:mi>v</m:mi>
                  <m:mi>in</m:mi>
                </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:mi>out</m:mi>
		  </m:msub>
		</m:ci>
              </m:apply>
	    </m:apply>
	  </m:math> 
	  or
	  <m:math>
	    <m:apply>
	      <m:eq/>
              <m:ci>
                <m:msub>
                  <m:mi>i</m:mi>
                  <m:mi>out</m:mi>
                </m:msub>
              </m:ci>
              <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>2</m:mn>
		  </m:msub>
		</m:ci>
              </m:apply>
	    </m:apply>
	  </m:math>.
	  This result does not depend on the resistor
	  <m:math>
	    <m:ci>
	      <m:msub>
		<m:mi>R</m:mi>
		<m:mn>1</m:mn>
	      </m:msub>
	    </m:ci>
	  </m:math>, 
	  which  means  that  we  simply  have  a  resistor
	  (<m:math>
	    <m:ci>
	      <m:msub>
		<m:mi>R</m:mi>
		<m:mn>2</m:mn>
	      </m:msub>
	    </m:ci>
	  </m:math>)
	  across a voltage source. The two-resistor circuit has no
	  apparent use.
	</para>
      </solution>
    </exercise>


    <para id="para3lit2">
      This circuit highlights some important properties of parallel
      circuits. You can easily show that the parallel 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> 
      has the <emphasis>v-i</emphasis> relation of a resistor having
      resistance
      <m:math>
	<m:apply>
	  <m:eq/>
          <m:apply>
            <m:power/>
	    <m:apply>
	      <m:plus/>
	      <m:apply>
		<m:divide/>
		<m:cn>1</m:cn>
		<m:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mn>1</m:mn>
		  </m:msub>
		</m:ci>
	      </m:apply>
	      <m:apply>
		<m:divide/>
		<m:cn>1</m:cn>
		<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:minus/>
	      <m:cn>1</m:cn>
	    </m:apply>
          </m:apply>
          <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>.
      A shorthand notation for this quantity is  
      <m:math>
	<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:math>.  As the reciprocal of resistance is <cnxn document="m0012" target="res" strength="5">conductance</cnxn>,
      we can say that
      <emphasis>
	for a parallel combination of resistors, the equivalent
	conductance is the sum of the conductances
      </emphasis>.
    </para>

    <figure id="parallelR">
      <media type="image/png" src="parallelR.png"/>
    </figure>
    
    <para id="para4lit2">
      Similar to <cnxn target="voltdivide" strength="4">voltage
      divider</cnxn> for series resistances, we have <term>current
      divider</term> for parallel resistances.  The current through a
      resistor in parallel with another is the ratio of the
      conductance of the first to the sum of the conductances. Thus,
      for the depicted circuit,
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>
	    <m:msub>
	      <m:mi>i</m:mi>
	      <m:mn>2</m:mn>
	    </m:msub>
	  </m:ci>
	  <m:apply>
	    <m:times/>
	    <m:apply>
	      <m:divide/>
	      <m:ci>
		<m:msub>
		  <m:mi>G</m:mi>
		  <m:mn>2</m:mn>
		</m:msub>
	      </m:ci>
	      <m:apply>
		<m:plus/>
		<m:ci>
		  <m:msub>
		    <m:mi>G</m:mi>
		    <m:mn>1</m:mn>
		  </m:msub>
		</m:ci>
		<m:ci>
		  <m:msub>
		    <m:mi>G</m:mi>
		    <m:mn>2</m:mn>
		  </m:msub>
		</m:ci>
	      </m:apply>
	    </m:apply>
	    <m:ci>i</m:ci>
	  </m:apply>
	</m:apply>
      </m:math>.  Expressed in terms of resistances, current divider
      takes the form of the resistance of the
      <emphasis>other</emphasis> resistor divided by the sum of
      resistances:
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>
	    <m:msub>
	      <m:mi>i</m:mi>
	      <m:mn>2</m:mn>
	    </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>i</m:ci>
	  </m:apply>
	</m:apply>
      </m:math>.
    </para>
    
    <figure id="circuit14">
      <media type="image/png" src="circuit14.png"/>
    </figure>
    
    <figure id="attenscope">
      <media type="image/png" src="circuit4c.png"/>
      <caption>
	The <cnxn target="simplecircuitgwar" strength="5">simple
        attenuator circuit</cnxn> is attached to an oscilloscope's
        input.  The input-output relation for the above circuit
        without a load is: <m:math> <m:apply> <m:eq/> <m:ci> <m:msub>
        <m:mi>v</m:mi> <m:mi>out</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>.
      </caption>
    </figure>

    <para id="para1rub">
      Suppose we want to pass the output signal into a voltage
      measurement device, such as an oscilloscope or a voltmeter. In
      system-theory terms, we want to pass our circuit's output to a
      sink. For most applications, we can represent these measurement
      devices as a resistor, with the current passing through it
      driving the measurement device through some type of display.
      In circuits, a sink is called a <term>load</term>; thus, we
      describe a system-theoretic sink as 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>.
      Thus, we have a complete system built from a cascade of three
      systems: a source, a signal processing system (simple as it is),
      and a sink. 
    </para>
    
    <para id="para3fir2"> 
      We must analyze afresh how this revised circuit, shown in 
      <cnxn target="attenscope" strength="5"/>, 
      works. Rather than defining eight variables and solving for the
      current in the load resistor, let's take a hint from
      other analysis
      (<cnxn target="voltdivide" strength="5">series rules</cnxn>, 
      <cnxn target="para3lit2" strength="5">parallel rules</cnxn>). 
      Resistors  
      <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:mi>L</m:mi>
	  </m:msub>
	</m:ci>
      </m:math>
      are in a <term>parallel</term> configuration: The voltages
      across each resistor are the same while the currents are
      not. Because the voltages are the same, we can find the current
      through each from their <emphasis>v-i</emphasis> relations:
      <m:math>
	<m:apply>
	  <m:eq/>
          <m:ci>
            <m:msub>
              <m:mi>i</m:mi>
              <m:mn>2</m:mn>
            </m:msub>
          </m:ci>
          <m:apply>
            <m:divide/>
	    <m:ci>
	      <m:msub>
		<m:mi>v</m:mi>
		<m:mi>out</m:mi>
	      </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
      <m:math>
	<m:apply>
	  <m:eq/>
          <m:ci>
            <m:msub>
              <m:mi>i</m:mi>
              <m:mi>L</m:mi>
            </m:msub>
          </m:ci>
          <m:apply>
            <m:divide/>
	    <m:ci>
	      <m:msub>
		<m:mi>v</m:mi>
		<m:mi>out</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:math>.
      Considering the node where all three resistors join, KCL says
      that the sum of the three currents must equal zero. Said another
      way, the current entering the node through
      <m:math>
	<m:ci>
	  <m:msub>
	    <m:mi>R</m:mi>
	    <m:mn>1</m:mn>
	  </m:msub>
	</m:ci>
      </m:math>
      must equal the sum of the other two currents leaving the
      node. Therefore,
      <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:apply>
            <m:plus/>
	    <m:ci>
	      <m:msub>
		<m:mi>i</m:mi>
		<m:mn>2</m:mn>
	      </m:msub>
	    </m:ci>
	    <m:ci>
	      <m:msub>
		<m:mi>i</m:mi>
		<m:mi>L</m:mi>
	      </m:msub>
	    </m:ci>
          </m:apply>
	</m:apply>
      </m:math>,
      which means that  
      <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:apply>
	    <m:times/>
	    <m:ci>
	      <m:msub>
		<m:mi>v</m:mi>
		<m:mi>out</m:mi>
	      </m:msub>
	    </m:ci>
	    <m:apply>
	      <m:plus/>
	      <m:apply>
		<m:divide/>
		<m:cn>1</m:cn>
		<m:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mn>2</m:mn>
		  </m:msub>
		</m:ci>
	      </m:apply>
	      <m:apply>
		<m:divide/>
		<m:cn>1</m:cn>
		<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:apply>
      </m:math>.
    </para>

    <para id="para4"> 
      Let  
      <m:math>
	<m:ci>
	  <m:msub>
	    <m:mi>R</m:mi>
	    <m:mi>eq</m:mi>
	  </m:msub>
	</m:ci>
      </m:math>
      denote the equivalent resistance of the parallel 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:mi>L</m:mi>
	  </m:msub>
	</m:ci>
      </m:math>. Using 
      <m:math>
	<m:ci>
	  <m:msub>
	    <m:mi>R</m:mi>
	    <m:mn>1</m:mn>
	  </m:msub>
	</m:ci>
      </m:math>'s  
      <emphasis>v-i</emphasis> relation, the voltage across it is  
      <m:math>
	<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: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>v</m:mi>
		  <m:mi>out</m:mi>
		</m:msub>
	      </m:ci>
	    </m:apply>
	    <m:ci>
	      <m:msub>
		<m:mi>R</m:mi>
		<m:mi>eq</m:mi>
	      </m:msub>
	    </m:ci>
	  </m:apply>
	</m:apply>
      </m:math>.
      The KVL equation written around the leftmost loop has  
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>
	    <m:msub>
	      <m:mi>v</m:mi>
	      <m:mi>in</m:mi>
	    </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:mi>out</m:mi>
	      </m:msub>
	    </m:ci>
	  </m:apply>
	</m:apply>
      </m:math>;
      substituting for  
      <m:math>
	<m:ci>
	  <m:msub>
	    <m:mi>v</m:mi>
	    <m:mn>1</m:mn>
	  </m:msub>
	</m:ci>
      </m:math>,
      we find      
                                                   
	<m:math display="block">
	  <m:apply>
	    <m:eq/>
	    <m:ci>
	      <m:msub>
		<m:mi>v</m:mi>
		<m:mi>in</m:mi>
	      </m:msub>
	    </m:ci>
	    <m:apply>
	      <m:times/>
	      <m:ci>
		<m:msub>
		  <m:mi>v</m:mi>
		  <m:mi>out</m:mi>
		</m:msub>
	      </m:ci>
	      <m:apply>
		<m:plus/>
		<m:apply>
		  <m:divide/>
		  <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:mi>eq</m:mi>
		    </m:msub>
		  </m:ci>
		</m:apply>
		<m:cn>1</m:cn>
	      </m:apply>
	    </m:apply>
	  </m:apply>
	</m:math>

      or

	<m:math display="block">
	  <m:apply>
	    <m:eq/>
	    <m:apply>
	      <m:divide/>
	      <m:ci>
		<m:msub>
		  <m:mi>v</m:mi>
		  <m:mi>out</m:mi>
		</m:msub>
	      </m:ci>
	      <m:ci>
		<m:msub>
		  <m:mi>v</m:mi>
		  <m:mi>in</m:mi>
		</m:msub>
	      </m:ci>
	    </m:apply>
	    <m:apply>
	      <m:divide/>
	      <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mi>eq</m:mi>
		</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:mi>eq</m:mi>
		  </m:msub>
		</m:ci>
	      </m:apply>
	    </m:apply>
	  </m:apply>
	</m:math>
    </para>   

    <para id="para5"> 
      Thus, we have the input-output relationship for our entire
      system having the form of voltage divider, but it does
      <emphasis>not</emphasis> equal the input-output relation of the
      circuit without the voltage measurement device. We can not
      measure voltages reliably unless the measurement device has
      little effect on what we are trying to measure. We should look
      more carefully to determine if any values for the load
      resistance would lessen its impact on the circuit. Comparing the
      input-output relations before and after, what we need is
      <m:math>
	<m:apply>
	  <m:approx/>
	  <m:ci>
	    <m:msub>
	      <m:mi>R</m:mi>
	      <m:mi>eq</m:mi>
	    </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:math>. 
      As  
      <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:power/>
	    <m:apply>
	      <m:plus/>
	      <m:apply>
		<m:divide/>
		<m:cn>1</m:cn>
		<m:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mn>2</m:mn>
		  </m:msub>
		</m:ci>
	      </m:apply>
	      <m:apply>
		<m:divide/>
		<m:cn>1</m:cn>
		<m:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mi>L</m:mi>
		  </m:msub>
		</m:ci>
	      </m:apply>
	    </m:apply>
	    <m:apply>
	      <m:minus/>
	      <m:cn>1</m:cn>
	    </m:apply>
	  </m:apply>
	</m:apply>
      </m:math>, 
      the approximation would apply if 
      <m:math>
	<m:apply>
	  <m:ci><m:mo>≫</m:mo></m:ci>
	  <m:apply>
	    <m:divide/>
	    <m:cn>1</m:cn>
	    <m:ci>
	      <m:msub>
		<m:mi>R</m:mi>
		<m:mn>2</m:mn>
	      </m:msub>
	    </m:ci>
	  </m:apply>
	  <m:apply>
	    <m:divide/>
	    <m:cn>1</m:cn>
	    <m:ci>
	      <m:msub>
		<m:mi>R</m:mi>
		<m:mi>L</m:mi>
	      </m:msub>
	    </m:ci>
	  </m:apply>
	</m:apply>
      </m:math>
      or   
      <m:math>
	<m:apply>
	  <m:ci><m:mo>≪</m:mo></m:ci>
	  <m:ci>
	    <m:msub>
	      <m:mi>R</m:mi>
	      <m:mi>2</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:math>.
      This is the condition we seek:  

      <note type="Voltage measurement">
	Voltage measurement devices must have large resistances
	compared with that of the resistor across which the voltage is
	to be measured.
      </note>
    </para>

    <exercise id="exer1lit2">
      <problem>
	<para id="exer1alit2">
	  Let's be more precise: How much larger would a load
	  resistance need to be to affect the input-output relation by
	  less than 10%? by less than 1%?
	</para>
      </problem>

      <solution>
	<para id="exer1blit2"> 
	  <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:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mn>2</m:mn>
		  </m:msub>
		</m:ci>
		<m:apply>
		  <m:plus/>
		  <m:cn>1</m:cn>
		  <m:apply>
		    <m:divide/>
		    <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:mi>L</m:mi>        
		      </m:msub>
		    </m:ci>
		  </m:apply>
		</m:apply>
              </m:apply>
	    </m:apply>
	  </m:math>.
	  Thus, a 10% change means that the ratio  
	  <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:ci>
                <m:msub>
                  <m:mi>R</m:mi>
                  <m:mi>L</m:mi>
                </m:msub>
              </m:ci>
	    </m:apply>
	  </m:math>
	  must be less than 0.1. A 1% change means that  
	  <m:math>
	    <m:apply>
	      <m:lt/>
              <m:apply>
                <m:divide/>
		<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:mi>L</m:mi>
		  </m:msub>
		</m:ci>
              </m:apply>
              <m:cn>0.01</m:cn>
	    </m:apply>
	  </m:math>.
	</para>
      </solution>
    </exercise>


    <example id="ex1">

      <figure id="fg2">
	<media type="image/png" src="circuit15.png"/>
      </figure>

      <para id="para6"> 
	We want to find the total resistance of the example
	circuit. To apply the series and parallel combination rules,
	it is best to first determine the circuit's structure: What is
	in series with what and what is in parallel with what at both
	small- and large-scale views. We have
	<m:math>
	  <m:ci>
	    <m:msub>
	      <m:mi>R</m:mi>
	      <m:mn>2</m:mn>
	    </m:msub>
	  </m:ci>
	</m:math> 
	in parallel with  
	<m:math>
	  <m:ci>
	    <m:msub>
	      <m:mi>R</m:mi>
	      <m:mn>3</m:mn>
	    </m:msub>
	  </m:ci>
	</m:math>;
	this combination is in series with  
	<m:math>
	  <m:ci>
	    <m:msub>
	      <m:mi>R</m:mi>
	      <m:mn>4</m:mn>
	    </m:msub>
	  </m:ci>
	</m:math>.
	This series combination is in parallel with 
	<m:math>
	  <m:ci>
	    <m:msub>
	      <m:mi>R</m:mi>
	      <m:mn>1</m:mn>
	    </m:msub>
	  </m:ci>
	</m:math>.
	Note that in determining this structure, we started
	<emphasis>away </emphasis>from the terminals, and worked
	toward them. In most cases, this approach works well; try it
	first. The total resistance expression mimics the structure:

	  <m:math display="block">
	    <m:apply>
	      <m:eq/>
	      <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mi>T</m:mi>
		</m:msub>
	      </m:ci>
	      <m:apply>
		<m:times/>
		<!-- not actual relation used for || (parallel) -->
		<m:ci>
		  <m:msub>
		    <m:mi>R</m:mi>
		    <m:mn>1</m:mn>
		  </m:msub>
		</m:ci>
		<m:ci><m:mo>∥</m:mo></m:ci>
		<m:apply>
		  <m:plus/>
		  <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: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:ci>
		    <m:msub>
		      <m:mi>R</m:mi>
		      <m:mn>4</m:mn>
		    </m:msub>
		  </m:ci>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:math>

	  <m:math display="block">
	    <m:apply>
	      <m:eq/>
	      <m:ci>
		<m:msub>
		  <m:mi>R</m:mi>
		  <m:mi>T</m:mi>
		</m:msub>
	      </m:ci>
	      <m:apply>
		<m:divide/>
		<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>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: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:ci>
		      <m:msub>
			<m:mi>R</m:mi>
			<m:mn>4</m:mn>
		      </m:msub>
		    </m:ci>
		  </m:apply>
		  <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>3</m:mn>
		      </m:msub>
		    </m:ci>
		    <m:ci>
		      <m:msub>
			<m:mi>R</m:mi>
			<m:mn>4</m:mn>
		      </m:msub>
		    </m:ci>
		  </m:apply>
		</m:apply>
		<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>R</m:mi>
			<m:mn>2</m:mn>
		      </m:msub>
		    </m:ci>
		  </m:apply>
		  <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>3</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>R</m:mi>
			<m:mn>3</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>R</m:mi>
			<m:mn>4</m:mn>
		      </m:msub>
		    </m:ci>
		  </m:apply>
		  <m:apply>
		    <m:times/>
		    <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>4</m:mn>
		      </m:msub>
		    </m:ci>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:math>

	Such complicated expressions typify circuit "simplifications."
	A simple check for accuracy is the units: Each component of
	the numerator should have the same units (here
	<m:math>
	  <m:apply>
	    <m:power/>
	    <m:ci>Ω</m:ci>
	    <m:cn>3</m:cn>
	  </m:apply>
	</m:math>)
	as well as in the denominator
	(<m:math>
	  <m:apply>
	    <m:power/>
	    <m:ci>Ω</m:ci>
	    <m:cn>2</m:cn>
	  </m:apply>
	</m:math>).
	The entire expression is to have units of resistance; thus,
	the ratio of the numerator's and denominator's units should be
	ohms. Checking units does not guarantee accuracy, but can
	catch many errors.      
      </para>
    </example>        
    

    <para id="para7">
      Another valuable lesson emerges from this example concerning the
      difference between cascading systems and cascading circuits. In
      system theory, systems can be cascaded without changing the
      input-output relation of intermediate systems. In cascading
      circuits, this ideal is rarely true <emphasis>unless</emphasis>
      the circuits are so
      <emphasis>designed</emphasis>. Design is in the hands of the
      engineer; he or she must recognize what have come to be known as
      loading effects. In our simple circuit, you might think that
      making the resistance
      <m:math>
	<m:ci>
	  <m:msub>
	    <m:mi>R</m:mi>
	    <m:mi>L</m:mi>
	  </m:msub>
	</m:ci>
      </m:math>
      large enough would do the trick. Because the resistors
      <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>
      can have virtually any value, you can never make the resistance
      of your voltage measurement device big enough. Said another way,
      <emphasis>a circuit cannot be designed in isolation that will
      work in cascade with all other circuits</emphasis>. Electrical
      engineers deal with this situation through the notion of
      <emphasis>specifications</emphasis>: Under what conditions will
      the circuit perform as designed? Thus, you will find that
      oscilloscopes and voltmeters have their internal resistances
      clearly stated, enabling you to determine whether the voltage
      you measure closely equals what was present before they were
      attached to your circuit. Furthermore, since our resistor
      circuit functions as an attenuator, with the attenuation (a
      fancy word for gains less than one) depending only on the ratio
      of the two resistor values
      <m:math>
	<m:apply>
	  <m:eq/>
	  <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:apply>
	    <m:power/>
	    <m:apply>
	      <m:plus/>
	      <m:cn>1</m:cn>
	      <m:apply>
		<m:divide/>
		<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:minus/>
	      <m:cn>1</m:cn>
	    </m:apply>
	  </m:apply>
	</m:apply>
      </m:math>, 
      we can select <emphasis>any</emphasis> values for the two
      resistances we want to achieve the desired attenuation. The
      designer of this circuit must thus specify not only what the
      attenuation is, but also the resistance values employed so that
      integrators—people who put systems together from
      component systems—can combine systems together and have a
      chance of the combination working.
    </para>

    <para id="one"> 
      <cnxn target="combo" strength="5"/> 
      summarizes the series and parallel combination results.  These
      results are easy to remember and very useful. Keep in mind that
      for series combinations, voltage and resistance are the key
      quantities, while for parallel combinations current and
      conductance are more important. In series combinations, the
      currents through each element are the same; in parallel ones,
      the voltages are the same.      
    </para>


    <figure orient="horizontal" id="combo">
      <name>series and parallel combination rules</name>
      <subfigure id="comboa">
	<name>series combination rule</name>
	<media type="image/png" src="seriesex.png"/>
	<caption>
	  <m:math>
	    <m:apply>
	      <m:eq/>
	      <m:ci>
		<m:msub>
		  <m:ci>R</m:ci>
		  <m:ci>T</m:ci> 
		</m:msub>
	      </m:ci>    
	      <m:apply>
		<m:sum/>
                <m:bvar>
                  <m:ci>n</m:ci>
                </m:bvar>
                <m:lowlimit>
                  <m:cn>1</m:cn>
                </m:lowlimit>
                <m:uplimit>
                  <m:ci>N</m:ci>
                </m:uplimit>
                <m:apply>
                  <m:ci>
                    <m:msub>
                      <m:ci>R</m:ci>
                      <m:ci>n</m:ci>
                    </m:msub>
                  </m:ci>
                </m:apply>
	      </m:apply>
	    </m:apply>
	  </m:math>  
	    
	  <m:math>
	    <m:apply>
	      <m:eq/>
	      <m:ci>
		<m:msub>
		  <m:ci>v</m:ci>
		  <m:ci>n</m:ci>
		</m:msub>
	      </m:ci> 
	      <m:apply>
		<m:times/>
                <m:apply>
                  <m:divide/>
		  <m:ci>
		    <m:msub>
		      <m:ci>R</m:ci>
		      <m:ci>n</m:ci>
		    </m:msub>
		  </m:ci>
		  <m:ci>
		    <m:msub>
		      <m:ci>R</m:ci>
		      <m:ci>T</m:ci> 
		    </m:msub>
		  </m:ci>
                </m:apply>
                <m:ci>v</m:ci>
	      </m:apply>
	    </m:apply>
	  </m:math>
	</caption>
      </subfigure>
      <subfigure id="combob">
	<name>parallel combination rule</name>  
	<media type="image/png" src="parallelex.png"/>
	<caption> 
	  <m:math>
	    <m:apply>
	      <m:eq/>
	      <m:ci>
		<m:msub>
		  <m:ci>G</m:ci>
		  <m:ci>T</m:ci> 
		</m:msub>    
	      </m:ci>
	      <m:apply>
		<m:sum/>
                <m:bvar>
                  <m:ci>n</m:ci>
                </m:bvar>
                <m:lowlimit>
                  <m:cn>1</m:cn>
                </m:lowlimit>
                <m:uplimit>
                  <m:ci>N</m:ci>
                </m:uplimit>
                <m:ci>
                  <m:msub>
                    <m:ci>G</m:ci>
                    <m:ci>n</m:ci>
                  </m:msub>
                </m:ci>
	      </m:apply>
	    </m:apply>
	  </m:math>  
	                                       
	  <m:math>
	    <m:apply>
	      <m:eq/>
	      <m:ci>
		<m:msub>
		  <m:ci>i</m:ci>
		  <m:ci>n</m:ci>
		</m:msub>
	      </m:ci> 
	      <m:apply>
		<m:times/>
                <m:apply>
                  <m:divide/>
		  <m:ci>
		    <m:msub>
		      <m:ci>G</m:ci>
		      <m:ci>n</m:ci>
		    </m:msub>
		  </m:ci>
		  <m:ci>
		    <m:msub>
		      <m:ci>G</m:ci>
		      <m:ci>T</m:ci> 
		    </m:msub>
		  </m:ci>
                </m:apply>
                <m:ci>i</m:ci>
	      </m:apply>
	    </m:apply>
	  </m:math>
	</caption>
      </subfigure>
      <caption> 
	Series and parallel combination rules.
      </caption>
    </figure>  

                                                              
    <exercise id="exer1ice2">
      <problem>
	<para id="exer1aice2">
	  Contrast a series combination of resistors with a parallel
	  one. Which variable (voltage or current) is the same for
	  each and which differs? What are the equivalent resistances?
	  When resistors are placed in series, is the equivalent
	  resistance bigger, in between, or smaller than the component
	  resistances? What is this relationship for a parallel
	  combination?
	</para>
      </problem>

      <solution>
	<para id="exer1bice2">
	  In a series combination of resistors, the current is the
	  same in each; in a parallel combination, the voltage is the
	  same. For a series combination, the equivalent resistance is
	  the sum of the resistances, which will be larger than any
	  component resistor's value; for a parallel combination, the
	  equivalent conductance is the sum of the component
	  conductances, which is larger than any component
	  conductance.  The equivalent resistance is therefore smaller
	  than any component resistance.
	</para>                                                             
      </solution>
    </exercise>    
  </content>  
</document>
