It really helped me, made my concepts clear and solved many doubts. For example, the circuit in Figure 7.2.9 could be solved using nodal analysis by converting the voltage source and the associated resistance into a current source. . Each iteration creates a new equation. Is atmospheric nitrogen chemically necessary for life? Also decide on the reference node (usually ground). Determine the load current. Nodal Analysis is based on the application of the Kirchhoff's Current Law (KCL). Figure 3 Equivalent circuit for Example 1, At the single independent node, Equation 1 becomes, ${{I}_{5}}+{{I}_{1}}+{{I}_{10}}={{I}_{gen}}+{{I}_{bat}}$. This is not a problem. The current source is directing current into node a, and we will assume the current flows away from a towards node b and c. . The resistor in question would have to be much smaller than any surrounding resistors in order to have minimal impact on the results. Thank you @nav, Sorry for that typo; It's V_E = I_6 * R_6 as you said. Solution These terms will all be negative. Deploy nodal analysis method to solve the circuit and find the power of the dependent source. Continue with Recommended Cookies, Home Electrical Circuits Nodal Analysis with Solved Examples Label the directions of branch currents flowing in the circuit with respect to reference node V3. Solution 1) Identify all nodes in the circuit. I_5 = 180 I_x All quantities are known except for \(V_b\) and thus it is easily found with a little more algebra. We then write current summation equations at each node (except for ground). Find \(V_a\) and \(V_b\) for the circuit of Figure 7.2.13 Identify all nodes in the circuit. Solving 'n-1' equations all the nodes voltages can be obtained. Node #2: II. For current sources, a more direct approach is possible. We also know that \(V_a V_b = 60\) volts. //--> This circuit has four current summing nodes; ground and nodes \(a\), \(b\), and \(c\). This method requires that we solve the simultaneous equations that result when Kirchoff's current law (KCL) is applied at various nodes in an electric circuit. Node : . Also, it uses only current sources and no voltage sources. : Circuit for Example 7.2.3 Therefore, instead of writing equations using three nodes, we shall instead refer to node \(c\) in reference to node \(a\), that is we shall write \(V_a\) 8 V wherever we need \(V_c\). The currents are then described by their Ohm's law equivalents: \[\text{Node } a: I_1 = \frac{V_a V_b}{R_3} + \frac{V_a}{R_1} \nonumber \], \[\text{Node } b: I_2 = \frac{V_a V_b}{R_3} +\frac{V_b}{R_2} \nonumber \], \[\text{Node } a: I_1 = \left( \frac{1}{R_1} + \frac{1}{R_3} \right) V_a \left( \frac{1}{R_3} \right) V_b \nonumber \], \[\text{Node } b: I_2 = \left( \frac{1}{R_3} \right) V_a + \left( \frac{1}{R_3} + \frac{1}{R_2} \right) V_b \nonumber \]. //--> Yaz is here. : Basic dual voltage source circuit with currents and nodes defined. As \(V_c\) is 8 volts less than \(V_a\), then \(V_c = 2\) volts. google_ad_slot = "7352692842"; Use MathJax to format equations. Sample Problem: Mesh/Node (Supernode) 10:38. Note the value for E1 is expressed in terms of an unknown value. . on Nodal Analysis Dependent Current Source, Nodal Analysis Dependent Voltage Source (5-Nodes), Nodal Analysis Dependent Voltage Source. View more posts, thanks a lot !! . Whenever we're doing nodal analysis, we always start with identifying the nodes in the circuit, and we know that we have a node at the upper left-hand side of the circuit, we'll call it node 1. Here is the trap: in the converted circuit, although \(R_1\) still connects to node \(a\), the other end no longer connects to the voltage source. Figure 7.2.13 . Figure 7.2.9 Finally, simplify the constants and coefficients, and the first expression is complete: \[1.2 A = 130mS V_a 30mS V_b \nonumber \]. Similarly, we consider all resistor currents to flow out of a node. We also know that \(V_a V_b = E\) from the original circuit. Supernode analysis is nodal analysis when a voltage source exists between two nonreference nodes. Solution: Step 1: To select a reference node and label the node voltages Step 2: Apply Kirchhoff's current law to all the nodes Node 1: I s 2 + V 1 V 2 R 3 + V 1 V 3 R 1 = 0 6 V 1 V 2 5 V 3 = 5 (eq.1) Node 2: Fewer nodes means fewer equations and a quicker solution. google_ad_height = 280; For node \(a\), assuming \(I_1\) exits as drawn: \[I_1 = 1A \frac{V_a}{4 \Omega } \frac{V_aV_b}{20 \Omega } \nonumber \], \[I_1 = 1 A \frac{12.857V}{4 \Omega } \frac{12.857V(47.143 V)}{20 \Omega } \nonumber \], \[I_1 =5.2143A \text{ (negative exit means it's entering)} \nonumber \]. Now, we need to find the voltage across the dependent current source and the current passing through it. Good luck:), Nodal analysis for current controlled current source, Speeding software innovation with low-code/no-code tools, Tips and tricks for succeeding as a developer emigrating to Japan (Ep. 1 and Eq.2 leads to /* solved problems top on single pages */ google_ad_slot = "7352692842"; /* circuits middle 468*15 */ } Call the number of nodes . We can also try to eliminate as many equations from them as possible using any info we have; however in the end, we need to be left with. We will build the first expression piece by piece. : Circuit for Example 7.2.2 Solve the circuit with nodal analysis and find and . Did you find apk for android? Figure 7.2.1 "@type": "ListItem", Points \(a\) and \(c\) are places where components connect, but they are not summing nodes, so we can ignore them for now. its more than a lectures hall. For example, if we define the currents through R7 and R8 as I7 and I8, both defined in the "top-to-bottom" direction as the circuit is drawn, then we can easily find. Stack Exchange network consists of 182 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. },{ V_A - V_B = I_7 R_7 \\ google_ad_height = 15; Nodal Analysis with Current Source From the above circuit, there are 3 nodes. The circuit has 4 nodes: Therefore, . : Circuit modified for supernode analysis. Writing this in terms of Ohm's law we have: \[I_x I_y = \frac{1}{R_1} V_a + \frac{1}{R_3} V_b \nonumber \]. . , Solution Hi! Okay here I go. The fixed current sources are: Next, we find all of the resistors connected to node \(b\) and write them as conductances, the group being multiplied by node voltage \(b\): \[800mA 300 mA = \left( \frac{1}{25 \Omega } + \frac{1}{50 \Omega } + \frac{1}{100 \Omega } \right) V_b \dots \nonumber \]. We are thus restating Kirchhoffs current law in the form: At any independent node, the algebraic sum of the resistor currents leaving the node equals the algebraic sum of the source currents entering the node. "item": The 20V voltage source reduces the unknowns by 1. Notify me of follow-up comments by email. Once again, there will be as many equations as node voltages. Consider the circuit of Figure 7.2.2 Consequently, we require only one. Solution a) Choose a reference node, label the voltages: b) Apply KCL to nodes: Node #1: Node #2: Node #3: . Nodal analysis circuit uses the Kirchhoff's current law (KCL) For the 'n' nodes (including reference node) there will be 'n-1' independent nodal voltage equations Solving all the equations will grant us the nodal voltages value The number of nodes (except non-reference nodes) is equal to the number of the nodal voltage equation we can get. Or, we can go directly to the format of Equation 2, $\begin{align} & \left( {{G}_{5}}+{{G}_{1}}+{{G}_{10}} \right)V={{I}_{gen}}+{{I}_{bat}} \\& V=\frac{80+128}{5+1+10}=13V \\& {{I}_{L}}={{G}_{L}}{{V}_{L}}=13V\times 1S=13A \\\end{align}$. : Basic dual current source circuit with currents and nodes defined. Thanks for contributing an answer to Electrical Engineering Stack Exchange! Label it with reference (ground) symbol. For the other node conductances, determine the conductances between the node under inspection and these other nodes. Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. In this example there is only one other node, node \(b\), and thus only one iteration. $\begin{matrix} {{I}_{R1}}+{{I}_{R2}}+\cdots ={{I}_{S1}}+{{I}_{S2}}+\cdots & {} & \left( 1 \right) \\\end{matrix}$if(typeof ez_ad_units!='undefined'){ez_ad_units.push([[468,60],'electricala2z_com-box-3','ezslot_4',105,'0','0'])};__ez_fad_position('div-gpt-ad-electricala2z_com-box-3-0'); The next step is to replace the resistor currents using I = VG. We now write a current summation equation for each summing node, except for the reference node. This would remove the node and simplify creating the system of equations, but not alter the remainder of the circuit. 9/16/2019 Nodal Analysis - Supernode - Solved Problems 2/12Solution 1) Identify all nodes in the circuit. Step 3 Write nodal equations at all the principal nodes except the reference node. { Without that resistance, it becomes impossible to create an expression for the current passing through the source using the general method, and impossible to convert the voltage source into a current source in order to use the inspection method. google_ad_client = "ca-pub-5562743950047265"; The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. "@id": "https://electricala2z.com/electrical-circuits/nodal-analysis-solved-examples/", On occasion you may come across a circuit like the one shown in Figure 7.2.11 For example, a portion of a network might consist of two parallel resistors which are in series with a third resistor. Typically, the reference node is ground, although it does not have to be. By solving the system of equations, If voltage sources exist, they must be converted to current sources before proceeding. Call the number of nodes . Let's verify that these values are correct. Node has a voltage source connected to. google_ad_width = 336; 1). "url": "https://electricala2z.com/category/electrical-circuits/", The Kirchhoffs current-law equations then have the format. Next, find all of the resistors connected to the node of interest and write them as a sum of conductances on the other side of the equals sign, the group being multiplied by this node's voltage (e.g., \(V_1\)). Solution I. As \(V_b\) is higher than the 6 volt source, our assumed current direction for the 4 k\( \Omega \) resistor was incorrect; we assumed right to left but it is in fact left to right, flowing from 11.379 volts to 6 volts. If there are dependent sources in the circuit, write down equations that express their values in terms of other node voltages. We start with the observation that \(V_c = V_a 8\) V. In other words, \(V_c\) is locked to \(V_a\) and if we find one of them, we can determine the other. Determine the current "i" through the resistor R_1 for the circuit in the figure 3.3. b) Apply KCL to nodes: As shown in Figure 7.2.14 ), write down their behaviour (for instance, ohm's law for a resistance, i = c dV/dt for a capacitance and so on), At this point, we'll have a handful of equations with us. Verify that the circuit uses only current sources with resistors and no voltage sources. Solving 'n-1' equations all the nodes voltages can be obtained. An alternative to the basic supernode technique is to simply describe one node voltage in terms of another at the outset. google_ad_width = 468; I know SRS, however I am not sure of how to apply it in this problem. In MNA, you would write KCL at each node (except ground) and consider the currents in the voltage sources as additional unknowns. I_x + I_5 = I_6 Consider the currents entering and exiting this combined or super node. Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. When the migration is complete, you will access your Teams at stackoverflowteams.com, and they will no longer appear in the left sidebar on stackoverflow.com. shown with conductances instead of resistances. Furthermore, we don't have to deal with the painful Thevenin equivalent/Super-mesh workarounds here On the downside, I'm not quite sure if this approach works with every possible circuit (so far I haven't seen any where it fails), so any negative feedback on this part is welcome :). Figure 7.2.2 One possible way out of this quandary is to simply add a very small resistor in series with it so that a source conversion is possible. 30 Homework Statement: Solve for current I0 using nodal analysis. Contents Application of Nodal Analysis Application of nodal analysis requires taking a node in the circuit as a reference or datum node. This time the voltage source is left in. Determine \(V_a\) and \(V_b\) in the circuit of Figure 7.2.6 Nodal analysis significantly reduces the computational work required by reducing the set of equations to be solved in a particular circuit or network. \[800mA 300 mA = \left( \frac{1}{50 \Omega } + \frac{1}{100 \Omega } \right) V_a+ \left( \frac{1}{25 \Omega } + \frac{1}{50 \Omega } + \frac{1}{100 \Omega } \right) V_b \nonumber \]. Node is defined as a junction or joining point of two or more component terminals. google_ad_client = "ca-pub-5562743950047265"; Save my name, email, and website in this browser for the next time I comment. Select a reference node. Equation 2 illustrates the standard format for writing a nodal equation. We will examine two variations; a general version that can be used with both voltage and current sources, and a second somewhat quicker version that can be used with circuits only driven by current sources. Sample Problem: Mesh Analysis (Depend Sources) 1 6:48. google_ad_slot = "4163172045"; At first glance, Figure 2 appears to show six junction points, but there are only two because the same voltage appears across all parallel branches. There will be as many equations as there are nodes, less the reference node. Legal. that has a voltage source without a series resistance associated with it. \[1.5A = (0.2S +0.25S) V_a (0.25S) V_b (0) V_c \nonumber \], \[1.5A = 0.45S V_a 0.25S V_b 0 V_c \nonumber \]. /* bottom single */ Using in this KCL equation introduces an unnecessary unknown to the equations set. 4 = 0 + 3v1 + 3v3 Eq 1. : Circuit of Example 7.2.5 Noting that \(\frac{V_a = E_1}\) and \(\frac{V_c = E_2}\), with a little algebra this can be reduced to a series of products of conductances and voltages: \[\left( \frac{1}{R_1} \right) E_1 + \left( \frac{1}{R_2} \right) E_2 = \left( \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right) V_b \nonumber \]. Thank you. \[\left( {{G}_{2}}+{{G}_{3}} \right){{V}_{2}}-{{G}_{3}}{{V}_{1}}=-{{I}_{S2}}\]if(typeof ez_ad_units!='undefined'){ez_ad_units.push([[300,250],'electricala2z_com-box-4','ezslot_5',108,'0','0'])};__ez_fad_position('div-gpt-ad-electricala2z_com-box-4-0'); Nodal analysis is particularly useful for networks where a common portion of the network is fed from several sources in parallel. From this positive term, we must subtract a term for the node voltage at every adjacent node connected by a resistor to the node in question. The only conductance common between nodes \(a\) and \(b\) is \(1/R_3\), yielding the \(V_b\) coefficient. 7: Nodal and Mesh Analysis, Dependent Sources, DC Electrical Circuit Analysis - A Practical Approach (Fiore), { "7.1:_Introduction" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "7.2:_Nodal_Analysis" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "7.3:_Mesh_Analysis" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "7.4:_Dependent_Sources" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "7.5:_Summary" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "7.6:_Exercises" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()" }, { "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "01:_Fundamentals" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "02:_Basic_Quantities" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "03:_Series_Resistive_Circuits" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "04:_Parallel_Resistive_Circuits" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "05:_Series-Parallel_Resistive_Circuits" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "06:_Analysis_Theorems_and_Techniques" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "07:_Nodal_and_Mesh_Analysis_Dependent_Sources" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "08:_Capacitors" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "09:_Inductors" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "10:_Magnetic_Circuits_and_Transformers" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()" }, [ "article:topic", "license:ccbyncsa", "showtoc:no", "authorname:jmfiore", "licenseversion:40", "source@http://www.dissidents.com/resources/DCElectricalCircuitAnalysis.pdf" ], https://eng.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Feng.libretexts.org%2FBookshelves%2FElectrical_Engineering%2FElectronics%2FDC_Electrical_Circuit_Analysis_-_A_Practical_Approach_(Fiore)%2F07%253A_Nodal_and_Mesh_Analysis_Dependent_Sources%2F7.2%253A_Nodal_Analysis, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), Converting Sources and Other Simplifications, source@http://www.dissidents.com/resources/DCElectricalCircuitAnalysis.pdf, status page at https://status.libretexts.org. Therefore, \[\frac{V_a V_b}{R_1} + \frac{V_c V_b}{R_2} = \frac{V_b}{R_3} \nonumber \]. google_ad_slot = "8829426048"; In general use, it might be considered a universal solution technique as there are no practical circuit configurations that it cannot handle. Thus, this three node circuit will only need two equations. google_ad_client = "ca-pub-5562743950047265"; if(typeof ez_ad_units!='undefined'){ez_ad_units.push([[300,250],'electricala2z_com-medrectangle-4','ezslot_3',107,'0','0'])};__ez_fad_position('div-gpt-ad-electricala2z_com-medrectangle-4-0'); Figure 1 circuit diagram for writing nodal equations. What do you do in order to drag out lectures? If you can figure out how a "(resistor + current_source in series) parallel to (volt_source + resistor in parallel)" works, I think you'll be able to play with any circuit. In this video, we have discussed the Concept of Dependent Sources & solved problems of the topic Nodal Analysis with Dependent Voltage Source and Current Sou. For comparison sake, this circuit was solved in Chapter 6 using source conversions and also using superposition. All examples and problems contain detailed analysis of various circuits, and are solved using a recipe approach, providing a code that motivates students to decode and apply to real-life engineering scenarios Covers the basic topics of resistors, voltage and current sources, capacitors and inductors, Ohms and Kirchhoffs Laws, nodal and mesh . We can perform a KCL summation at node \(a\) and see if it balances. Now we repeat the entire process for the next equation. and match the calculated values perfectly. with currents labeled and using equivalent conductances in place of the resistances. Generally to solve this type of circuits we find a relation to express Ix in terms of node voltages / resistance. Entering is deemed positive while exiting is deemed negative. only use KVL KCL MESH analysis, Node analysis, and source. Multiply those conductances by the other associated node voltages and subtract those products from the expression built so far. Otherwise, sensor nodes use the cooperative combining strategies scheme to send the data from the source to the destination or sink node. Fixed :), @niko: Frankly speaking, analog circuits are real sources of pain. RACE-SM solved the single relay node issues by using the sink mobility scheme. Figure 7.2.12 Example 1: Using Nodal method, find the current through resistor r2 (Figure 1). To view the purposes they believe they have legitimate interest for, or to object to this data processing use the vendor list link below. Node 1 corresponds to \(V_b\), node 2 corresponds to \(V_a\) and node 3 corresponds to \(V_c\). (Eq. /* middle on single pages */