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<TITLE>Thevenin Equivalents</TITLE>
A cell behaves like a perfect voltage 
            source in series with a resistor. A <b>perfect voltage source</b> 
            is one whose voltage does not drop under load. The 
            value of this &quot;resistor&quot; is called<b> source impedance</b>, 
            (a.k.a. <i>internal resistance, Thevenin resistance </i>or<i> </i> 
            <i>R</i><sub>Th</sub>). </p>
          <p align="left">The voltage of the perfect 
            source is equal to the <b>no-load voltage</b> (also called <i>open 
            circuit</i> <i>voltage</i> or <i>Thevenin voltage,</i> <i>V</i><sub><font size="-1">Th</font></sub>). 
            <i>V</i><sub><font size="-1">out</font></sub><i>=V</i><sub><font size="-1">Th</font></sub> 
            when <i>I</i>=0 </p>
          <p align="left"><img src="thevenin-schematic.gif" width="300" height="211"></p>
          <p align="left">Source impedance increases 
            and no-load voltage decreases as the cell discharges. The 
            no-load voltage of a cell, battery or power supply sitting on your 
            bench, not connected to a load, is easy to measure directly. Source 
            impedance, on the other hand, is measured under operating conditions.</p>
          <p align="left">Q: The no-load voltage 
            of a cell is 3V. The output voltage <br>&nbsp;&nbsp;&nbsp;&nbsp;at 750mA is 1.9V. What is the 
            internal resistance?</p>
          <p align="left">A: The voltage drop is 
            3V-1.9V=1.1V<br>&nbsp;&nbsp;&nbsp;&nbsp;<i>R=V/I</i>=1.1V/.75A=<b>1.47 ohm</b></p>
          <p align="left">Q: What's the load resistance?</p>
          <p align="left">A: <i>R=V/I</i>=1.9V/.75A=<b>2.53 ohm</b>. 
            </p>
          <p align="left">To calculate source impedance, 
            connect a resistor to the cell and measure the current:<br>
            If load resistance is known, <i>R</i><sub><font size="-1">Th</font></sub><i> 
            = (V</i><sub><font size="-1">Th</font></sub><i>/I)-R</i><sub><font size="-1">load</font></sub><br>
            If output voltage is known, <i>R</i><sub><font size="-1">Th</font></sub><i> 
            = (V</i><sub><font size="-1">Th</font></sub><i>-V</i><sub><font size="-1">out</font></sub><i>)/I</i> 
                  
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