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| learn:courses:real-analog-chapter-3:start [2017/04/20 17:16] – [3. Introduction and Chapter Objectives] Martha | learn:courses:real-analog-chapter-3:start [2022/10/04 11:38] (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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| ==== After Completing this Chapter, You Should be Able to: ==== | ==== After Completing this Chapter, You Should be Able to: ==== | ||
| * Use nodal analysis techniques to analyze electrical circuits | * Use nodal analysis techniques to analyze electrical circuits | ||
| - | * Use mesh analysis techniques to analyze electrical circuits | + | * Use mesh analysis techniques to analyze electrical circuits |
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| Node b: | Node b: | ||
| - | $$-i_1+i_3+i_4=0 | + | $$-i_1+i_3+i_4=0 |
| Node c: | Node c: | ||
| - | $$-i_2-i_3+i_5=0 | + | $$-i_2-i_3+i_5=0 |
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| ==== Step 5: Use Ohm's Law to Write the Equations From Step 4 in Terms of Voltages ==== | ==== Step 5: Use Ohm's Law to Write the Equations From Step 4 in Terms of Voltages ==== | ||
| - | The currents defined in Step 4 can be written in terms of the node voltages defined previously. For example, from Fig. 3.7: $i_1= \frac{V_S-V-b}{R_1}$, | + | The currents defined in Step 4 can be written in terms of the node voltages defined previously. For example, from Fig. 3.7: $i_1= \frac{V_S-V-b}{R_1}$, |
| $$-\frac{V_s-V_b}{R_1} + \frac{V_b-V_c}{R_3} + \frac{V_b-0}{R_4} = 0 $$ | $$-\frac{V_s-V_b}{R_1} + \frac{V_b-V_c}{R_3} + \frac{V_b-0}{R_4} = 0 $$ | ||
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| So the KCL equation for node b becomes: | So the KCL equation for node b becomes: | ||
| - | $$ \left( \frac{1}{R_1} + \frac{1}{R_3} + \frac{1}{R_4} \right) V_b - \frac{1}{R_3} V_c = \frac{1}{R_1}V_S | + | $$ \left( \frac{1}{R_1} + \frac{1}{R_3} + \frac{1}{R_4} \right) V_b - \frac{1}{R_3} V_c = \frac{1}{R_1}V_S |
| Likewise, the KCL equation for node c can be written as: | Likewise, the KCL equation for node c can be written as: | ||
| - | $$-\frac{1}{R_3} V_b + \left( \frac{1}{R_3} + \frac{1}{R_2} + \frac{1}{R_5} \right) V_c = \frac{1}{R_2} V_S | + | $$-\frac{1}{R_3} V_b + \left( \frac{1}{R_3} + \frac{1}{R_2} + \frac{1}{R_5} \right) V_c = \frac{1}{R_2} V_S |
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| ==== Double-checking Results ==== | ==== Double-checking Results ==== | ||
| If the circuit being analyzed contains only __independent sources__, and the sign convention used in KCL equations is the same as used above (__currents leaving nodes are assumed positive__), | If the circuit being analyzed contains only __independent sources__, and the sign convention used in KCL equations is the same as used above (__currents leaving nodes are assumed positive__), | ||
| - | * The term multiplying the voltage at that node will be the sum of the conductances connected to that node. For the example above, the term multiplying $V_b$ in the equation for node b is $\frac{1}{R_1} + \frac{1}{R_3} + \frac{1}{R_4}$ while the term multiplying $V_c$ nin the equation for node c is $\frac{1}{R_3} + \frac{1}{R_2} + \frac{1}{R_5}$. | + | * The term multiplying the voltage at that node will be the sum of the conductances connected to that node. For the example above, the term multiplying $V_b$ in the equation for node b is $\frac{1}{R_1} + \frac{1}{R_3} + \frac{1}{R_4}$ while the term multiplying $V_c$ in the equation for node c is $\frac{1}{R_3} + \frac{1}{R_2} + \frac{1}{R_5}$. |
| * The term multiplying the voltages adjacent to the node will be the negative of the conductance connecting the two nodes. For the example above, the term multiplying $V_c$ in the equation for node b is $-\frac{1}{R_3}$, | * The term multiplying the voltages adjacent to the node will be the negative of the conductance connecting the two nodes. For the example above, the term multiplying $V_c$ in the equation for node b is $-\frac{1}{R_3}$, | ||
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| $$V=V_A - V_B = -2V$$ | $$V=V_A - V_B = -2V$$ | ||
| - | Several comments should be made relative to the above example: | + | Several comments should be made relative to the above example:\\ |
| - | 1. Steps 4 and 5 (applying KCL at each independent node and using Ohm's law to write these equations in terms of voltages) have been combined into a single step. This approach is fairly common, and can provide a significant savings in time. | + | **1.** Steps 4 and 5 (applying KCL at each independent node and using Ohm's law to write these equations in terms of voltages) have been combined into a single step. This approach is fairly common, and can provide a significant savings in time. |
| - | 2. There may be a perceived inconsistency between the two node equations, in the assumption of positive current direction in the 1Ω resistor. In the equation for node A, the current is apparently assumed to be positive from node A to node B, as shown below: | + | **2.** There may be a perceived inconsistency between the two node equations, in the assumption of positive current direction in the 1Ω resistor. In the equation for node A, the current is apparently assumed to be positive from node A to node B, as shown below: |
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| - | This definition leads to the corresponding term in the equation for node B becoming: $\frac{V_B - V_A}{1}$/ | + | This definition leads to the corresponding term in the equation for node B becoming: $\frac{V_B - V_A}{1}$. |
| The above inconsistency in sign is, however, insignificant. Suppose that we had assumed (consistently with the equation for node A) that the direction of positive current for the node B equation is from Node A to B. Then, the corresponding term in the equation for node B would have been: $-\frac{V_A - V_B}{1}$ (note that a negative sign has been applied to this term to accommodate our assumption that __currents flowing into nodes are negative__). This is equal to $\frac{V_B - V_A}{1}$, which is exactly what our original result was. | The above inconsistency in sign is, however, insignificant. Suppose that we had assumed (consistently with the equation for node A) that the direction of positive current for the node B equation is from Node A to B. Then, the corresponding term in the equation for node B would have been: $-\frac{V_A - V_B}{1}$ (note that a negative sign has been applied to this term to accommodate our assumption that __currents flowing into nodes are negative__). This is equal to $\frac{V_B - V_A}{1}$, which is exactly what our original result was. | ||
| - | 3. The current source appears __directly__ in the nodal equations. | + | **3.** The current source appears __directly__ in the nodal equations. |
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| ==== Supernodes ==== | ==== Supernodes ==== | ||
| - | In the previous examples, we identified __dependent__ nodes and determined __constrained__ voltages. Kirchhoff' | + | In the previous examples, we identified __dependent__ nodes and determined __constrained__ voltages. Kirchhoff' |
| Example: for the circuit below, determine the voltage difference, V, across the 2mA source. | Example: for the circuit below, determine the voltage difference, V, across the 2mA source. | ||
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| * Use Ohm's law to write KCL at each independent node, in terms of the node voltages. This will result in N equations in N unknowns, where N is the number of node voltages. Independent " | * Use Ohm's law to write KCL at each independent node, in terms of the node voltages. This will result in N equations in N unknowns, where N is the number of node voltages. Independent " | ||
| * Solve the equations of step 4 to determine the node voltages. | * Solve the equations of step 4 to determine the node voltages. | ||
| - | * Use the node voltages to determine any other esired | + | * Use the node voltages to determine any other desired |
| * Modifications to the above approach are allowed. For example, it is not necessary to define supernodes in step 4 above. Once can define unknown voltages at either terminal of a voltage source and write KCL at each of these nodes. However, the unknown current through the voltage source must be accounted for when writing KCL - this introduces an additional unknown into the governing equations. This added unknown requires an additional equation. This equation is obtained by explicitly writing a constraint equation relating the voltages at the two terminals of the voltage source. | * Modifications to the above approach are allowed. For example, it is not necessary to define supernodes in step 4 above. Once can define unknown voltages at either terminal of a voltage source and write KCL at each of these nodes. However, the unknown current through the voltage source must be accounted for when writing KCL - this introduces an additional unknown into the governing equations. This added unknown requires an additional equation. This equation is obtained by explicitly writing a constraint equation relating the voltages at the two terminals of the voltage source. | ||
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| + | From the above figure, the voltage drops across the resistor $R_1$ can then be determined as: | ||
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| + | $$V_1 = R_1 i_1$$ | ||
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| + | Since only mesh current $i_1$ passes through the resistor $R_1$. Likewise, the voltage drop for the resistor $R_4$ is | ||
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| + | Since mesh currents $i_1$ and $i_4$ both pass through $R_4$ and the current $i_2$ is in the opposite direction to our assumed polarity for the voltage $V_4$. | ||
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| + | Using the above expressions for $V_1$ and $V_4$, we can write KVL for the first mesh loop as: | ||
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| + | $$-V_S + R_1 i_1 + R_4 (i_1 - i_2) =0$$ | ||
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| + | When we write KVL for the mesh current $i_2$ in our example, we choose voltage polarities for resistors $R_4$, $R_2$, and $R_5$ as shown in the figure below – these polarities agree with the passive sign convention for voltages relative to the direction of the mesh current $i_2$. | ||
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| Using the above sign conventions, | Using the above sign conventions, | ||
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| - | Define three mesh currents for each of the three meshes in the above circuit and define unknown voltages $V_1$ abd $V_3$ across the two current sources as shown below. | + | Define three mesh currents for each of the three meshes in the above circuit and define unknown voltages $V_1$ and $V_3$ across the two current sources as shown below. |
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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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