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| ====== Real Analog: Chapter 3 ====== | ====== Real Analog: Chapter 3 ====== | ||
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| + | * Lecture Material: | ||
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| + | * Chapter 3 Video: | ||
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| + | <-- | ||
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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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| Circuit reduction, it should be noted, is not fundamentally different from direct application of Ohm's and Kirchhoff' | Circuit reduction, it should be noted, is not fundamentally different from direct application of Ohm's and Kirchhoff' | ||
| - | In Chapter 1, we saw that direct application of Ohm's law and Kirchhoff' | + | In Chapter 1, we saw that direct application of Ohm's law and Kirchhoff' |
| In cases where circuit reduction is not feasible, approaches are still available to reduce the total number of unknowns in the system. //Nodal analysis// and //mesh analysis// are two of these. Nodal and mesh analysis approaches still rely upon application of Ohm's law and Kirchhoff' | In cases where circuit reduction is not feasible, approaches are still available to reduce the total number of unknowns in the system. //Nodal analysis// and //mesh analysis// are two of these. Nodal and mesh analysis approaches still rely upon application of Ohm's law and 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 |
| {{ : | {{ : | ||
| ==== 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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| This consists of two equations with three unknowns. The equation governing the dependent current source provides the third equation. Writing the controlling current in terms of independent voltages results in: | This consists of two equations with three unknowns. The equation governing the dependent current source provides the third equation. Writing the controlling current in terms of independent voltages results in: | ||
| - | $$I_X= \frac{V_A-0}{5\Omega} | + | $$I_X= \frac{V_A-0}{5\Omega}$$ |
| ---- | ---- | ||
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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: | ||
| + | |||
| + | $$V_1 = R_1 i_1$$ | ||
| + | |||
| + | Since only mesh current $i_1$ passes through the resistor $R_1$. Likewise, the voltage drop for the resistor $R_4$ is | ||
| + | |||
| + | Since mesh currents $i_1$ and $i_4$ both pass through $R_4$ and the current $i_2$ is in the opposite direction to our assumed polarity for the voltage $V_4$. | ||
| + | |||
| + | Using the above expressions for $V_1$ and $V_4$, we can write KVL for the first mesh loop as: | ||
| + | |||
| + | $$-V_S + R_1 i_1 + R_4 (i_1 - i_2) =0$$ | ||
| + | |||
| + | 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$. | ||
| + | |||
| + | {{ : | ||
| 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. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ===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: | ||
| + | |||
| + | $$-8V + 2\Omega \left( i_1 + 1A - 2A \right) + 4\Omega \left( i_1 - 2A \right) + 10V + 6\Omega \left( i_1 \right) = 0$$ | ||
| + | |||
| + | === Step 4: Solve the system of equations to determine the mesh currents of the circuit === | ||
| + | Solving the above equation results in $i_1 = 0.667A$. The current through the $4\Omega$ resistor is then, accounting for the 2A constrained loop passing through the resistor, $i = i_1 -2A = -1.333A$. | ||
| + | |||
| + | ==== Example 3.6: Version 2 ==== | ||
| + | In this version, we choose an alternate set of constraint loops. The alternate set of loops is shown below; all constraint loops still pass through only one current source, and retain the magnitude and direction of the source current. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Now, writing KVL for the single mesh results in: | ||
| + | |||
| + | $$-8V + 2\Omega \left( i_1 + 1A \right) + 4\Omega \cdot i_1 + 10V + 6\Omega \left( i_1 +2A \right) = 0$$ | ||
| + | |||
| + | Solving for the mesh current results in $i_1 = -1.333A$; note that this result is __different__ than previously. However, we determine the current through the $4\Omega$ resistor as $i=i_1 = -1.333A$, which __is__ the same result as previously. | ||
| + | |||
| + | ===Note=== | ||
| + | Choice of alternate constrained loops may change the values obtained for the mesh currents. The currents through the circuit elements, however, do not vary with choice of constrained loops. | ||
| + | |||
| + | ==== Example 3.6: Version 3 ==== | ||
| + | In this version, we choose yet another set of constrained loops. These loops are shown below. Again, each loop passes through one current source and retains that source' | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | KVL around the mesh loop results in: | ||
| + | |||
| + | $$-8V + 2\Omega \cdot i_1 + 4\Omega \left( i_1 - 1A \right) + 10V + 6\Omega \left( i_1 - 1A + 2A \right) = 0$$ | ||
| + | |||
| + | Which results in $i_1 =-0.333A$. Again, this is different from the result from our first two approaches. However, the current through the $4\Omega$ resistor is $i=i_1 - 1A = -1.333A$, which __is__ the same result as previously. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Dependent Sources ==== | ||
| + | As with nodal analysis, the presence of dependent sources does not significantly alter the overall mesh analysis approach. The primary difference is simply the addition of the additional equations necessary to describe the dependent sources. We discuss the analysis with dependent sources in the context of the following examples. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 3.7 ==== | ||
| + | Determine the voltage V in the circuit below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Shorting both of the voltage sources in the circuit above results in two mesh circuits. These are shown in the figure below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Writing KVL around the two mesh loops results in: | ||
| + | |||
| + | $$-2V + 2\Omega \cdot i_1 + 3\Omega \left( i_1-i_2 \right) = 0 $$ | ||
| + | |||
| + | $$2I_X + 3\Omega \left( i_2 - i_1 \right) + 4\Omega \cdot i_2 = 0$$ | ||
| + | |||
| + | we have two equations and three unknowns. We need an additional equation to solve the system of equations. The third equation is obtained by writing the dependent source' | ||
| + | |||
| + | $$I_X = i_1$$ | ||
| + | |||
| + | The above three equations can be solved to obtain $i_1 = 0.4375 \Omega$ and $i_2 = 0.0625 \Omega$. The desired voltage $V=4i_2 = 0.25V$. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 3.8 ==== | ||
| + | Write mesh equations for the circuit shown below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Mesh loops and constraint loops are identified as shown below: | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Writing KVL for the two mesh loops results in: | ||
| + | |||
| + | $$4\Omega \cdot i_1 + 2\Omega \left(i_1 - 3V_X \right) + 12V = 0$$ | ||
| + | |||
| + | $$-12V + 3\Omega \left(i_2-3V_X \right) + 5\Omega \cdot i_2 = 0$$ | ||
| + | |||
| + | Writing the controlling voltage $V_X$ in terms of the mesh currents results in: | ||
| + | |||
| + | $$V_X = 5\Omega \cdot i_2$$ | ||
| + | |||
| + | The above consist of three equations in three unknowns, which can be solved to determine the mesh currents. Any other desired circuit parameters can be determined from the mesh currents. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Section Summary ==== | ||
| + | * Basic steps in mesh analysis are: | ||
| + | * Identify mesh currents. This can be done by short-circuiting voltage sources, open-circuiting current sources, and identifying the enclosed, non-overlapping regions in the circuit. The perimeters of these areas are the circuit meshes. The mesh currents flow around the circuit meshes. | ||
| + | * Determine constrained loops. The approach in Step 1 will ensure that no mesh currents will pass through the current sources. The current source currents can be accounted for by defining constrained loops. Constrained loops are defined as loop currents which pass through the current sources. Constrained loops are identified by replacing the sources in the circuit schematic, and defining mesh currents which pass through the current sources; these mesh currents form the constrained loops and must match both the magnitude and direction of the current in the current sources. | ||
| + | * Use Ohm's law to write KVL around each mesh loop, in terms of the mesh currents. This results in N equations in N unknowns, where N is the number of mesh currents. Keep in mind that the voltage difference across each element must correspond to the voltage difference induced by __all__ the mesh currents which pass through that element. | ||
| + | * Solve the equations of Step 4 to determine the mesh currents. | ||
| + | * Use the mesh currents to determine any other desired voltages/ | ||
| + | * The constrained loops in Step 2 above are not unique. Their only requirement is that they must account for the currents through the current sources. | ||
| + | * Modifications to the above approach are allowed. For example, it is not necessary to define constrained loops in Step 3 above. One can define (unknown) mesh currents which pass through the current sources and write KVL for these additional mesh currents. However, the unknown voltage across the current source must be accounted for when writing KVL - this introduces an additional unknown into the governing equations. This added unknown requires an additional equation which is obtained by explicitly writing a constraint equation equating algebraic sum of the mesh currents passing through a current source to the current provided by the source. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Exercises ==== | ||
| + | 1. Use mesh analysis to write a set of equations from which you can find $I_1$, the current through the $12\Omega$ resistor. Do not solve the equations. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | 2. Use mesh analysis to find the current $I$ flowing through the $10\Omega$ resistor in the circuit below. Compare your result to your solution to exercise 2 of section 3.2. | ||
| + | |||
| + | {{ : | ||
| + | [[{}/ | ||
| + | [[{}/ | ||