106 lines
6.7 KiB
HTML
106 lines
6.7 KiB
HTML
<script type="text/javascript">
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$(document).ready(function() {
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$("#r1_slider").slider({
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value: 1, min: 1, max: 10, step: 1, slide: schematic.component_slider,
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schematic: "ctrls", component: "R1", property: "r", analysis: "dc",
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})
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$("#r2_slider").slider({
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value: 1, min: 1, max: 10, step: 1, slide: schematic.component_slider,
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schematic: "ctrls", component: "R2", property: "r", analysis: "dc",
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})
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$("#r3_slider").slider({
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value: 1, min: 1, max: 10, step: 1, slide: schematic.component_slider,
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schematic: "ctrls", component: "R3", property: "r", analysis: "dc",
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})
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$("#r4_slider").slider({
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value: 1, min: 1, max: 10, step: 1, slide: schematic.component_slider,
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schematic: "ctrls", component: "R4", property: "r", analysis: "dc",
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})
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$("#slider").slider(); });
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</script>
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<b>Lab 2A: Superposition Experiment</b>
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<br><br><i>Note: This part of the lab is just to develop your intuition about
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superposition. There are no responses that need to be checked.</i>
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<br/><br/>Circuits with multiple sources can be hard to analyze as-is. For example, what is the voltage
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between the two terminals on the right of Figure 1?
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<center>
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<input width="425" type="hidden" height="150" id="schematic1" parts="" analyses="" class="schematic ctrls" name="test2" value="[["w",[160,64,184,64]],["w",[160,16,184,16]],["w",[64,16,112,16]],["w",[112,64,88,64]],["w",[64,64,88,64]],["g",[88,64,0],{},["0"]],["w",[112,64,160,64]],["w",[16,64,64,64]],["r",[160,16,0],{"name":"R4","r":"1"},["1","0"]],["r",[160,16,1],{"name":"R3","r":"1"},["1","2"]],["i",[112,64,6],{"name":"","value":"6A"},["0","2"]],["r",[64,16,0],{"name":"R2","r":"1"},["2","0"]],["r",[64,16,1],{"name":"R1","r":"1"},["2","3"]],["v",[16,16,0],{"name":"","value":"8V"},["3","0"]],["view",-24,0,2]]"/>
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Figure 1. Example multi-source circuit
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</center>
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<br/><br/>We can use superposition to make the analysis much easier.
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The circuit in Figure 1 can be decomposed into two separate
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subcircuits: one involving only the voltage source and one involving only the
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current source. We'll analyze each circuit separately and combine the
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results using superposition. Recall that to decompose a circuit for
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analysis, we'll pick each source in turn and set all the other sources
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to zero (i.e., voltage sources become short circuits and current
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sources become open circuits). The circuit above has two sources, so
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the decomposition produces two subcircuits, as shown in Figure 2.
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<center>
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<table><tr><td>
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<input style="display:inline;" width="425" type="hidden" height="150" id="schematic2" parts="" analyses="" class="schematic ctrls" name="test2" value="[["w",[160,64,184,64]],["w",[160,16,184,16]],["w",[64,16,112,16]],["w",[112,64,88,64]],["w",[64,64,88,64]],["g",[88,64,0],{},["0"]],["w",[112,64,160,64]],["w",[16,64,64,64]],["r",[160,16,0],{"name":"R4","r":"1"},["1","0"]],["r",[160,16,1],{"name":"R3","r":"1"},["1","2"]],["r",[64,16,0],{"name":"R2","r":"1"},["2","0"]],["r",[64,16,1],{"name":"R1","r":"1"},["2","3"]],["v",[16,16,0],{"name":"","value":"8V"},["3","0"]],["view",-24,0,2]]"/>
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(a) Subcircuit for analyzing contribution of voltage source
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</td><td>
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<input width="425" type="hidden" height="150" id="schematic3" parts="" analyses="" class="schematic ctrls" name="test2" value="[["w",[16,16,16,64]],["w",[160,64,184,64]],["w",[160,16,184,16]],["w",[64,16,112,16]],["w",[112,64,88,64]],["w",[64,64,88,64]],["g",[88,64,0],{},["0"]],["w",[112,64,160,64]],["w",[16,64,64,64]],["r",[160,16,0],{"name":"R4","r":"1"},["1","0"]],["r",[160,16,1],{"name":"R3","r":"1"},["1","2"]],["i",[112,64,6],{"name":"","value":"6A"},["0","2"]],["r",[64,16,0],{"name":"R2","r":"1"},["2","0"]],["r",[64,16,1],{"name":"R1","r":"1"},["2","3"]],["view",-24,0,2]]"/>
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(b) Subcircuit for analyzing contribution of current source
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</td></tr></table>
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<br>Figure 2. Decomposition of Figure 1 into subcircuits
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</center>
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<br/>Let's use the DC analysis capability of the schematic tool to see superposition
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in action. The sliders below control the resistances of R1, R2, R3 and R4 in all
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the diagrams. As you move the sliders, the schematic tool will adjust the appropriate
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resistance, perform a DC analysis and display the node voltages on the diagrams. Here's
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what you want to observe as you play with the sliders:
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<ul style="margin-left:2em;margin-top:1em;margin-right:2em;margin-bottom:1em;">
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<i>The voltage for a node in Figure 1 is the sum of the voltages for
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that node in Figures 2(a) and 2(b), just as predicted by
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superposition. (Note that due to round-off in the display of the
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voltages, the sum of the displayed voltages in Figure 2 may only be within
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.01 of the voltages displayed in Figure 1.)</i>
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</ul>
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<br>
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<center>
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<table><tr valign="top">
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<td>
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<table>
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<tr valign="top">
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<td>R1</td>
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<td>
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<div id="r1_slider" style="width:200px; height:10px; margin-left:15px"></div>
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</td>
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</tr>
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<tr valign="top">
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<td>R2</td>
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<td>
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<div id="r2_slider" style="width:200px; height:10px; margin-left:15px; margin-top:10px;"></div>
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</td>
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</tr>
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<tr valign="top">
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<td>R3</td>
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<td>
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<div id="r3_slider" style="width:200px; height:10px; margin-left:15px; margin-top:10px;"></div>
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</td>
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</tr>
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<tr valign="top">
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<td>R4</td>
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<td>
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<div id="r4_slider" style="width:200px; height:10px; margin-left:15px; margin-top:10px;"></div>
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</td>
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</tr>
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</table>
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</td></tr></table>
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</center>
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