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| learn:courses:real-analog-chapter-3:start [2017/04/20 21:43] – [Alternate Approach to Constraint Loops: Constraint Equations] Martha | learn:courses:real-analog-chapter-3:start [2023/02/09 02:47] (current) – external edit 127.0.0.1 | ||
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| ====== Real Analog: Chapter 3 ====== | ====== Real Analog: Chapter 3 ====== | ||
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| - | ====== 3. Introduction and Chapter Objectives | + | ===== 3. Introduction and Chapter Objectives ===== |
| - | In Chapters [[https:// | + | In Chapters [[/ |
| * Ohm's Law | * Ohm's Law | ||
| * Kirchhoff' | * Kirchhoff' | ||
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| - | === //Step 1//: Define mesh loops === | + | === Step 1: Define mesh loops === |
| Replacing the two current sources with open circuits and the two voltage sources with short circuits results in a single mesh current, $i_1$, as shown below. | Replacing the two current sources with open circuits and the two voltage sources with short circuits results in a single mesh current, $i_1$, as shown below. | ||
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| - | === //Step 2//: Constrained loops, version 1 === | + | === Step 2: Constrained loops, version 1 === |
| Initially, we choose the constrained loops shown below. Note that each loop passes through only one source and has the magnitude and direction imposed by the source. | Initially, we choose the constrained loops shown below. Note that each loop passes through only one source and has the magnitude and direction imposed by the source. | ||
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| - | ===//Step 3//: Write KVL around the mesh loops === | + | ===Step 3: Write KVL around the mesh loops === |
| Our example has only one mesh current, so only one KVL equation is required. This equation is: | Our example has only one mesh current, so only one KVL equation is required. This equation is: | ||
| $$-8V + 2\Omega \left( i_1 + 1A - 2A \right) + 4\Omega \left( i_1 - 2A \right) + 10V + 6\Omega \left( i_1 \right) = 0$$ | $$-8V + 2\Omega \left( i_1 + 1A - 2A \right) + 4\Omega \left( i_1 - 2A \right) + 10V + 6\Omega \left( i_1 \right) = 0$$ | ||
| - | === //Step 4//: Solve the system of equations to determine the mesh currents of the circuit === | + | === Step 4: Solve the system of equations to determine the mesh currents of the circuit === |
| Solving the above equation results in $i_1 = 0.667A$. The current through the $4\Omega$ resistor is then, accounting for the 2A constrained loop passing through the resistor, $i = i_1 -2A = -1.333A$. | Solving the above equation results in $i_1 = 0.667A$. The current through the $4\Omega$ resistor is then, accounting for the 2A constrained loop passing through the resistor, $i = i_1 -2A = -1.333A$. | ||
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