Derivatives and Integrals: An Annotated Discourse

<worksheet label="worksheet-networks" top="3cm" bottom="100pt">
      <title>Networks Worksheet</title>

      <introduction>
        <title>Basic laws for electrical circuits</title>
        <p>
          This two-page worksheet was generously donated to the sample article by Virgil Pierce at a CuratedCourses workshop in August<nbsp />2018.
          It has default (skinny) left and right margins,
          but we have specified longer top and bottom margins,
          with the top being the larger of the two.
        </p>
        <theorem>
          <title>Ohms Law</title>
          <p>
            The current through a resistor is proportional to the ratio of the <em> Voltage </em>
            to the <em> Resistance </em>
            <me>
              I = \frac{V}{R}
            </me>
            Or for our purposes
            <me>
              I R = V
            </me>
          </p>
        </theorem>
        <theorem>
          <title>Kirchoffs Current Law</title>
          <p>
            The sum of the currents in a network meeting at a point is zero.
            <me>
              \sum_{k=1}^n I_k = 0
            </me>
          </p>
        </theorem>
        <example>
          <title>Kirchoff's Current Law</title>
          <p>
            For the circuit below <m> I_1 + I_2 = I_3 </m>.
          </p>
          <image xml:id="worksheet-kirchoff-law" width="40%">
<latex-image>
\begin{circuitikz}
\draw (0,0)
to[R, i=$I_1$, *-o](2,2);
\draw (0,0)
to[R, i=$I_2$, *-o](2,-2);
\draw (-3, 0)
to[R, i=$I_3$, o-*](0,0);
\end{circuitikz}
</latex-image>
          </image>
        </example>
        <theorem>
          <title>Kirchoffs Voltage Law</title>
          <p>
            The sum of the voltages around any closed circuit
            (or subcircuit)
            is zero.
            <me>
              \sum_{k=1}^n V_k = 0
            </me>
          </p>
        </theorem>
        <p>
          Kirchoffs Current Law and Kirkoffs Voltage Law combined with Ohms Law gives for any circuit of resistors and sources a linear system that may
          (or may not)
          determine the currents.
        </p>
      </introduction>
      <page>
      </page>
      <page>
      <sidebyside width="45%" margins="0%">
        <exercise workspace="1.5in">
          <statement>
            <p>
              For the simple network pictured,
              calculuate the amperage in each part of the network by setting up a system of linear equations for the amperages.
            </p>
            <image xml:id="worksheet-simple-network">
<latex-image>
\begin{circuitikz}
\draw (0,0)
to[V, v=$11\quad \mbox{volt}$](0, 3)
to[short](2,3)
to[R, R=$4\quad \mbox{ohm}$] (3, 3)
to[short](4,3)
to[R, R=$3\quad \mbox{ohm}$](4,0)
to[short](3,0)
to[R, R=$3\quad \mbox{ohm}$](2,0)
to[short](0,0);
\end{circuitikz}
</latex-image>
            </image>
          </statement>
        </exercise>
        <exercise workspace="2in">
          <statement>
            <p>
              Compare it with a parallel circuit network.
              Calculate the amperage in each part of the network by setting up a system of linear equations for the amperages.
            </p>
            <image xml:id="worksheet-parallel-circuit">
<latex-image>
\begin{circuitikz}
\draw(0,0)
to[V, v=$11\quad \mbox{volt}$](0, 3)
to[short](5,3)
to[R, R=$4\quad \mbox{ohm}$](6,3)
to[short](7,3)
to[R, R=$3\quad \mbox{ohm}$](7,0)
to[short](6,0)
to[R, R=$3\quad \mbox{ohm}$](5,0)
to[short](0,0);
\draw(4,0)
to[R, R=$3\quad \mbox{ohm}$](4,3);
\end{circuitikz}
</latex-image>
            </image>
          </statement>
        </exercise>
      </sidebyside>
      <exercise workspace="1in">
        <statement>
          <p>
            Now for a more complicated network.
            Calculate the amperage in each part of the network by setting up a system of linear equations for the amperages.
          </p>
          <image xml:id="worksheet-complicated-network">
<latex-image>
\begin{circuitikz}
\draw(0,0)
to[V, v=$11 \quad \mbox{volt}$](0, 3)
to[short](5,3)
to[R, R=$4 \quad \mbox{ohm}$](6,3)
to[short](9,3)
to[R, R=$4 \quad \mbox{ohm}$](10,3)
to[short](11,3)
to[R, R=$3 \quad \mbox{ohm}$](11, 0)
to[short](10,0)
to[R, R=$2\quad \mbox{ohm}$](9,0)
to[short](6,0)
to[R, R=$2\quad \mbox{ohm}$](5,0)
to[short](0,0);
\draw(4,0)
to[R, R=$3\quad \mbox{ohm}$](4,3);
\draw(8,0)
to[R, R=$1\quad \mbox{ohm}$](8,3);
\end{circuitikz}
</latex-image>
          </image>
        </statement>
      </exercise>
      </page>
      <page>
      <exercise workspace="3in">
        <statement>
          <p>
            Now generalize these ideas to a context outside of electrical circuits.
            Consider the network of streets given in the diagram
            (with one-way directions as indicated).
          </p>
          <image xml:id="worksheet-street-network" width="65%">
<latex-image>

\begin{tikzpicture}[&gt;=stealth]
\draw[-&gt;, very thick] (0,0) -- (10, 0) node[midway, below]{East Bound Winooski Ave};
\draw[&lt;-, very thick] (0, 1) -- (10, 1) node[midway, above]{West Bound Winooski Ave};
\draw[&lt;-, very thick] (0, 4) -- (10, 4) node[midway, above]{Shelburne St};
\draw[&lt;-, very thick] (1, -1) -- (1, 5) node[midway, above, sloped]{Willow};
\draw[-&gt;, very thick] (9, -1) -- (9, 5) node[midway, above, sloped]{Jay};
\end{tikzpicture}

</latex-image>
          </image>
          <p>
            A traffic engineer counts the hourly flow of cars into and out of this network at the entrances.
            They get (EB = East Bound; WB = West Bound):
          </p>
          <table>
            <title>Estimated hourly traffic flow for the road network</title>
            <tabular row-headers="yes" halign="center">
              <row header="yes">
                <cell />
                <cell>EB Winooski</cell>
                <cell>WB Winooski</cell>
                <cell>Shelburne St</cell>
                <cell>Willow</cell>
                <cell>Jay</cell>
              </row>
              <row>
                <cell halign="left">into</cell>
                <cell>50</cell>
                <cell>400</cell>
                <cell>0</cell>
                <cell>10</cell>
                <cell>50</cell>
              </row>
              <row>
                <cell halign="left">out of</cell>
                <cell>55</cell>
                <cell>390</cell>
                <cell>20</cell>
                <cell>15</cell>
                <cell>30</cell>
              </row>
            </tabular>
          </table>
          <p>
            Use a variable for each segment inside of the network and set up a system of linear equations restricting the flow.
            Solve the system.
            Note that you should not get a unique solution as traffic should be able to flow through the network in various ways.
          </p>
        </statement>
      </exercise>
      </page>
    </worksheet>