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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' | ||
| Line 8: | Line 37: | ||
| 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 |
| ---- | ---- | ||
| Line 114: | Line 143: | ||
| 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: $i_1= \frac{V_S-V-b}{R_1}$, |
| $$-\frac{V_s-V_b}{R_1} + \frac{V_b-V_c}{R_3} + \frac{V_b-0}{R_4} = 0 $$ | $$-\frac{V_s-V_b}{R_1} + \frac{V_b-V_c}{R_3} + \frac{V_b-0}{R_4} = 0 $$ | ||
| Line 129: | Line 158: | ||
| 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 |
| Line 140: | Line 169: | ||
| ==== 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 $-\frac{1}{R_3}$, | ||
| Line 178: | Line 207: | ||
| $$V=V_A - V_B = -2V$$ | $$V=V_A - V_B = -2V$$ | ||
| - | 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: |
| {{ : | {{ : | ||
| Line 187: | Line 216: | ||
| {{ : | {{ : | ||
| - | 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. |
| Line 204: | Line 233: | ||
| ==== 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. | ||
| Line 299: | Line 328: | ||
| Node A: | Node A: | ||
| - | $$2A+ \frac{V_A - 0V}{4\Omega} + \frac{(V_A - 2V) - V_B}{4\Omega} + \frac{(V_A - 2V} - V_C}{8\Omega} = 0 \Rightarrow 5V_A - 2V_B - V_C = -10$$ | + | $$2A+ \frac{V_A - 0V}{4 \Omega} + \frac{( V_A-2V) -V_B}{4 \Omega} + \frac{(V_A - 2V) - V_C}{8\Omega} = 0 \Rightarrow 5V_A - 2V_B - V_C = -10$$ |
| Node B: | Node B: | ||
| Line 329: | Line 358: | ||
| We now, as previously, write KCL for each independent node, taking into account the dependent voltage resulting from the presence of the supernode: | We now, as previously, write KCL for each independent node, taking into account the dependent voltage resulting from the presence of the supernode: | ||
| + | |||
| + | $$\frac{(V_A-2V_X)-0}{R_3} + \frac{V_A-0}{R_4} + \frac{V_A-V_B}{R_2}=0$$ | ||
| + | |||
| + | $$\frac{V_B-V_A}{R_2} + \frac{V_B-(V_A+2V_X)}{R_1} -I_S=0$$ | ||
| + | |||
| + | The above equations result in a system with two equations and three unknowns: $V_A$, $V_B$, and $V_X$ ($I_S$ is a known current). We now write any equations governing the dependent sources. Writing the controlling voltage in terms of the independent voltages results in: | ||
| + | |||
| + | $$V_X=V_A-V_B$$ | ||
| + | |||
| + | We now have three equations in three unknowns, which can be solved to determine the independent voltages, $V_A$ and $V_B$. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 3.4 ==== | ||
| + | Write the nodal equations for the circuit below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The reference node, independent voltages and dependent voltages ae shown on the figure below. A supernode, consisting of the 4V source and the nodes on either side of it, exists but is not shown explicitly on the figure. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Applying KCL for each independent node results in: | ||
| + | |||
| + | $$\frac{(V_A+4V)-3V}{2\Omega} + \frac{(V_A + 4V)-0}{4\Omega} + \frac{V_A-0}{5\Omega} + \frac{V_A-V_B}{3\Omega} = 0$$ | ||
| + | |||
| + | $$\frac{V_B-V_A}{3\Omega}-3I_X=0$$ | ||
| + | |||
| + | 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: | ||
| + | |||
| + | $$I_X= \frac{V_A-0}{5\Omega}$$ | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Section Summary ==== | ||
| + | * Basic steps in nodal analysis are: | ||
| + | * Define a reference node. All node voltages will be relative to this reference voltage. | ||
| + | * Identify independent nodes. This can be done by short-circuiting voltage sources, open-circuiting current sources, and identifying the remaining nodes in the circuit. The voltages at these nodes are the node voltages. | ||
| + | * Determine dependent voltages. This can be done by replacing the sources in the circuit schematic, and writing voltage constraints introduced by voltage sources. | ||
| + | * 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. | ||
| + | * Use the node voltages to determine any other desired voltages/ | ||
| + | * 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. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Exercises === | ||
| + | 1. Use nodal 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 nodal analysis to find the current I flowing through the $10\Omega$ resistor in the circuit below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ===== 3.3: Mesh Analysis ==== | ||
| + | In mesh analysis, we will define a set of mesh currents and use Ohm's law to write Kirchhoff' | ||
| + | |||
| + | Mesh analysis is appropriate for //planar circuits//. Planar circuits can be drawn in a single plane((Essentially, | ||
| + | |||
| + | The steps used in mesh analysis are provided below. The steps are illustrated in terms of the circuit of Fig. 3.8. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ==== Step 1: Define Mesh Currents ==== | ||
| + | In order to identify our mesh loops, we will turn off all sources, much like what we did in nodal analysis. To do this, we: | ||
| + | * Short-circuit all voltage sources. | ||
| + | * Open-circuit all current sources. | ||
| + | |||
| + | Once the sources have been turned off, the circuit can be divided into a number of __non-overlapping__ areas, each of which is completely enclosed by circuit elements. The circuit elements bounding each of these areas form the meshes of our circuit. The mesh currents flow around these meshes. Our example circuit has two meshes after removal of the sources, the resulting mesh currents are as shown in Fig. 3.9. | ||
| + | |||
| + | === Note === | ||
| + | We will always choose our mesh currents as flowing __clockwise__ around the meshes. This assumption is not fundamental to the application of mesh analysis, but it will result in a special form for the resulting equations which will later allow us to do some checking of our results. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ==== Step 2: Replace Sources and Identify Constrained Loops ==== | ||
| + | The presence of current sources in our circuit will result in the removal of some meshes during Step 1. We must now account for these meshes in our analysis by returning the sources to the circuit and identifying // | ||
| + | |||
| + | We have two rules for constrained loops: | ||
| + | - Each current must have __one and only one__ constrained loop passing through it. | ||
| + | - The __direction and magnitude__ of the constrained loop current must agree with the direction and magnitude of the source current. | ||
| + | |||
| + | For our example circuit, we choose our constrained loop as shown below. It should be noted that constrained loops can, if desired, cross our mesh loops - we have, however, chosen the constrained loop so that is does not overlap any of our mesh loops. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ==== Step 3: Write KVL Around the Mesh Loops ==== | ||
| + | We will apply Kirchhoff' | ||
| + | |||
| + | Note that more than one mesh current may pass through a circuit element. When determining voltage drops across individual elements, the __contributions from all mesh currents passing through that element must be included in the voltage drop__. | ||
| + | |||
| + | When we write KVL for a given mesh loop, we will base our sign convention for the voltage drops on the direction of the mesh current for that loop. | ||
| + | |||
| + | For example, when we write KVL for the mesh current $i_1$ in our example, we choose voltage polarities for resistors $R_1$ and $R_4$ as shown in the figure below - these polarities agree with the passive sign convention for voltages __relative to the direction of the mesh current $i_1$__. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | 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, | ||
| + | |||
| + | $$R_4 \left( i_2-i_1 \right) + R_2i_2 + R_5 \left( i_2 + I_S \right) = 0 $$ | ||
| + | |||
| + | Please note that the currents $i_2$ and $I_s$ are in the same direction in the resistor $R_5$, resulting in a summation of these currents in the term corresponding to the voltage drop across the resistor $R_5$. | ||
| + | |||
| + | === Notes: === | ||
| + | - Assumed sign conventions on voltage drops for a particular mesh loop are based on the assumed direction of that loop's mesh current. | ||
| + | - The current passing through an element is the __algebraic sum__ of all mesh and constraint currents passing through that element. This algebraic sum of currents is used to determine the voltage drop of the element. | ||
| + | |||
| + | ==== Step 4: Solve the System of Equations to Determine the Mesh Currents of the Circuit ==== | ||
| + | Step 3 will always result in N equations in N unknowns, where N is the number of mesh currents identified in Step 1. These equations can be solved for the mesh currents. Any other desired circuit parameters can be determined from the mesh currents. | ||
| + | |||
| + | The example below illustrates the above approach. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 3.5 ==== | ||
| + | In the circuit below, determine the voltage drop, $V$, across the 3Ω resistor. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Removing the sources results in a single mesh loop with mesh current $i_1$, as shown below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Replacing the sources and defining one constrained loop per source results in the loop definitions shown below (note that each constrained loop goes through only one source and that the amplitude and direction of the constrained currents agrees with source). | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Applying KVL around the loop $i_1$ and using Ohm's law to write voltage drops in terms of currents: | ||
| + | |||
| + | $$-3V+7\Omega \left( i_1 + 1A \right) + 3\Omega \left( i_1 + 1A + 3A \right) + 4\Omega \left( i_1 + 3A \right) = 0 \Rightarrow i_1 = -2A$$ | ||
| + | |||
| + | Thus, the current $i_1$ is 2A, in the opposite direction to that shown. The voltage across the $3\Omega$ resistor is $V = 3\Omega \left( i_1 + 3A + 1A \right) = 3\Omega \left( -2A +3A +1A \right) = 3 \left( 2A \right) = 6V$. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Alternate Approach to Constraint Loops: Constraint Equations ==== | ||
| + | In the above examples, the presence of current sources resulted in a reduced number of meshes. Constraint loops were then used to account for current sources. An alternate approach, in which we retain additional mesh currents and then apply // | ||
| + | |||
| + | Example: determine the voltage, $V$, in the circuit 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. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Applying KVL around the three mesh loops results in three equations with five unknowns: | ||
| + | |||
| + | $$V_1 + 3\Omega \cdot \left(i_1 - i_3 \right) + 7\Omega \cdot \left( i_1 - i_2 \right) = 0$$ | ||
| + | |||
| + | $$-3V + 7\Omega \cdot \left(i_2 - i_1 \right) + V_3 = 0$$ | ||
| + | |||
| + | $$-V_3 + 3\Omega \cdot \left(i_3 - i_1 \right) + 4\Omega \cdot i_3 = 0$$ | ||
| + | |||
| + | Two additional //constrain equations// are necessary. These can be determined by the requirement that the algebraic sum of the mesh currents passing through a current source must equal the current provided by the source. Thus, we obtain: | ||
| + | |||
| + | $$-i_2 + i_3 = 3A$$ | ||
| + | |||
| + | $$-i_1 = 1A$$ | ||
| + | |||
| + | Solving the five simultaneous equations above results in the same answer determined previously. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Clarification: | ||
| + | Previously, it was claimed that the choice of constraint loops is somewhat arbitrary. The requirements are that each source has only one constraint loop passing through it, and that the magnitude and direction of the constrained loop current be consistent with the source. Since constraint loops __can__ overlap other mesh loops without invalidating the mesh analysis approach, the choice of constraint loops is not unique. The examples below illustrate the effect of different choices of constraint loops on the analysis of a particular circuit. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ====Example 3.6: Version 1 ==== | ||
| + | Using mesh analysis, determine the current, $i$, through the 4Ω resistor. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | === 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. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | === 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. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ===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. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | |||
| + | [[{}/ | ||
| + | [[{}/ | ||