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EENG 2009 Part 5. Nodal Analysis 5.1 5.2 5.3 5.4 Nodes and Node Voltages Nodal Analysis By Example Supernodes Dependent Sources EENG 2009 FS 2006 Part 5: Nodal Analysis 146 5.1 Nodes and Node Voltages Nodes Recall that nodes are the connected segments of conductor that remain when we remove the circuit elements (which at present for us are either resistors or sources). Example 1 Identify the nodes for the given circuit. 16 9A 8 12 3A Solution: Redraw the circuit with the circuit elements removed and note the connected segments of conductors. We see that there are 3 nodes for this circuit: A single node! EENG 2009 FS 2006 Part 5: Nodal Analysis 147 Node Voltages A node voltage associated with a given node is defined to be the voltage difference between the given node and a reference node, which has been chosen from among the nodes. For a circuit with N essential nodes, there are N–1 node voltages. Once the set of node voltages is determined, all the other voltages and currents can be obtained in a straightforward manner. Example 2 Identify a reference node and corresponding node voltages for the given circuit (whose nodes we found in the previous example). 16 9A 8 12 EENG 2009 FS 2006 Part 5: Nodal Analysis 3A 148 Example 2 Solution: Using the results of Example 1, we draw the circuit with the nodes emphasized: 16 9A 8 12 3A Next choose the bottom node as the reference node, and designate the node voltages for the other two nodes as v1 and v2: v1 v2 16 node 1 9A 8 node 2 12 3A Reference node, ground, earth, “sea level” EENG 2009 FS 2006 Part 5: Nodal Analysis 149 Measurement of Branch and Node Voltages voltmeter 16 V + – v1 16 The voltage being measured is a branch voltage. v2 1 9A 8 2 12 voltmeter 64 V + – 3A voltmeter Node Voltages! 48 V + – The voltage across the 16- branch is not a node voltage. It is a branch voltage, and is actually the difference between the two node voltages v1 and v2. EENG 2009 FS 2006 Part 5: Nodal Analysis 150 Branch Voltages In Terms of Node Voltages Example 3 Find the relationships among the branch voltages and the node voltages. v1 + v12 – v 2 + 2 1 + v13 – 3 v23 – Solution: There are three branch voltages, v13 , v12 , and v23, and two node voltages, v1 and v2 (not counting v3) . The branch voltages v13 and v23 are clearly equal to the node voltages v1 and v2: v13 = v1 v23 = v2 The branch voltage v12 is a combination of the node voltages v1 and v2. EENG 2009 FS 2006 Part 5: Nodal Analysis 151 Solution (cont.): We can write the branch voltage v12 in terms of the node voltages v13 and v23 by applying KVL, as follows: + v1 + v12 – v2 2 + 1 v13 – v23 – 3 v12 = v13 – v23 = v1 – v2 Note the correlation with the order of the subscripts: The subscripts of v12 are 1 & 2 and the subscripts of the two node voltages being subtracted are 1 & 2. In general, vj node j vk node k vjk = vj – vk EENG 2009 FS 2006 Part 5: Nodal Analysis Memorize me! 152 Branch Currents In Terms of Node Voltages Example 4 Find the relationships among the branch current and the node voltages. Note that neither voltagepolarity markings nor a current reference arrow is shown on the diagram! R v1 node 1 v2 node 2 reference node Solution: The branch current flowing from node 1 to node 2 is: i12 = ( v1 – v2 ) / R The branch current flowing from node 2 to node 1 is: i21 = ( v2 – v1 ) / R In general: vj node j R vk node k ijk = ( vj – vk ) / R ikj = ( vk – vj ) / R Note that to avoid confusion, ijk, ikj, and their corresponding reference directions are not shown on the diagram. EENG 2009 FS 2006 Part 5: Nodal Analysis 153 5.2 Nodal Analysis By Example The rationale for nodal analysis is that once the node voltages are determined, all the other voltages and currents can be obtained in a simple manner. The reference node is chosen by the circuit analyst. In electronic circuits, we frequently choose the node to which lots of branches are connected. In power systems, we usually choose “ground” or “earth.” Look for the associated symbol on the circuit: Basic Procedure For Nodal Analysis: 1. Identify the nodes and the node voltages. 2. Write KCL at the proper nodes. (Here’s where we use the branch-current / node-voltage relationship we just developed.) 3. Solve for the node voltages. 4. Use the node voltages to solve for any other required quantities. EENG 2009 FS 2006 Part 5: Nodal Analysis 154 Example 1 Find the values of the node voltages. R2 is1 R1 R3 is2 Solution: Step 1. There are 3 nodes. Choose the bottom node as the reference node, and designate the other two nodes as v1 and v2. v2 v1 R2 node 2 node 1 is1 R1 R3 is2 node 3 With 3 nodes and one of them as the reference node, there are two nodal equations needed. EENG 2009 FS 2006 Part 5: Nodal Analysis 155 Solution (cont.): Step 2. Write KCL at nodes 1 and 2. v1 v2 R 2 node 1 is1 R1 node 2 R3 is2 KCL at node 1 (Summing the currents flowing out of node 1): v1 / R1 + (v1 – v2) / R2 – is1 = 0 At node 2: (v2 – v1) / R2 + v2 / R3 + is2 = 0 These are the equations to solve for v1 and v2. We can put them in orderly form (and discover a shortcut to writing them), as we show next. EENG 2009 FS 2006 Part 5: Nodal Analysis 156 Development of the Algorithmic Method for Writing Nodal- Analysis Equations Consider the previous circuit and its nodal equations: v1 is1 R1 R2 v2 is2 R3 v1 / R1 + (v1 – v2) / R2 – is1 = 0 (v2 – v1) / R2 + v2 / R3 + is2 = 0 Collect terms in v1 and v2 and put the independent-source terms on the RHS: (1 / R1 + 1 / R2) v1 – 1/ R2 v2 = is1 – 1/ R2 v1 + (1 / R2 + 1 / R3) v2 = – is2 EENG 2009 FS 2006 Part 5: Nodal Analysis 157 Now, using conductances: (G1 + G2) v1 – G2 v2 = is1 – G2 v1 + (G2 + G3) v2 = – is2 v1 R2 v2 is1 R1 R3 (1) (2) is2 Observe that the following algorithm is valid: In (1), the coefficient of v1 = (conductances connected to node 1) In (1), the coefficient of v2 = – (conductances between nodes 1 & 2) In (2), the coefficient of v2 = (conductances connected to node 2) In (2), the coefficient of v1 = – (conductances between nodes 1 & 2) In (1) and (2) the RHS = (current sources entering the node) EENG 2009 FS 2006 Part 5: Nodal Analysis 158 Example 2 Find v1 and v2. 16 + 9A + 8 3A V1 V2 - Solution: 16 v1 node 1 9A 12 v2 Electrically the same as the node 2 node 2 8 3A 12 KCL @ node 1: v1 / 8 + (v1 – v2) / 16 – 9 = 0 KCL @ node 2: (v2 – v1) / 16 + v2 /12 – 3 = 0 Collecting terms: (1/8 + 1/16) v1 – 1/16 v2 –1/16 v1 + (1/16 + 1/12) v2 = 9 = 3 The solution is v1 = 64 V, v2 = 48 V. EENG 2009 FS 2006 Part 5: Nodal Analysis 159 Solution (cont.) Note that these equations could have been written directly by using the algorithm! v1 9A 8 16 v2 3A 12 (1/8 + 1/16)v1 - 1/16 v2 = 9 –1/16 v1 + (1/16 + 1/12) v2 = 3 The TI-85 & TI-86 keystrokes for solving this equation are: 2nd SIMULT Number = 2 ENTER a1,1 = 8 –1 + 16 –1 ENTER a1,2 = – 16 –1 ENTER b1 = 9 ENTER a2,1 = – 16 –1 ENTER a2,2 = 16 –1 + 12 –1 ENTER b2 = 3 ENTER SOLVE x1 = 64.000 x2 = 48.000 EENG 2009 FS 2006 Part 5: Nodal Analysis 160 Example 3. Find the voltage differences across the sources. 4 5A 4 6A 2 3A Solution: 4 3A node 2 v2 v1 node 1 5A 6A 4 2 KCL @ nodes 1 and 2: (v1 – v2) / 4 + 5 + v1 / 2 – 3 = 0 (v2 – v1) / 4 – 5 – 6 + v2 / 4 = 0 Solution: v1 = – 4 V, v2 = 20 V EENG 2009 FS 2006 Part 5: Nodal Analysis 161 Example 3. (cont.) We still have to find the voltage differences across the three sources, as follows.* v2 = 20 V 5A 4 v1 = – 4 V 3A + v3A – + v5A –6 A 2 + v6A – 4 v3A = v1 = – 4 V v6A = v2 = 20 V v5A = v2 – v1 = 24 V * At this point it is up to us to choose the polarity markings for the sources, as none were specified. EENG 2009 FS 2006 Part 5: Nodal Analysis 162 12 Example 4. Find the current i and the power absorbed by the 6 resistor. 5A 6 i 4 18 6 Solution: v2 12 node 2 6 5A i 4 v3 node 3 node 1 v1 18 6 KCL @ 1: (v1 – v3)/4 + (v1 – v2)/12 + v1/6 = 0 KCL @ 2: (v2 – v1)/12 + (v2 – v3)/6 – 5 = 0 KCL @ 3: v3/18 + (v3 – v1)/4 + (v3 – v2)/6 = 0 Solving for the node voltages gives: v1 = 21 V, v2 = 45 V, v3 = 27 V EENG 2009 FS 2006 Part 5: Nodal Analysis 163 Solution (cont.): To calculate i we apply Ohm’s Law (which requires calculating the branch voltage across the 4 i 4 resistor): v3 v1 i = ( v3 – v1) / 4 = (27 – 21) / 4 = 3/2 A* * Now we know that the true current thru the 4 resistor really is flowing from left-to-right. At node 1 we wrote the expression for the current flowing from right to left, and at node 3 from left-to-right. To find the power absorbed by the 6 resistor: v2 = 45 V 6 12 p6 = (v2 – v3)2 / 6 4 5A v1 v3 = 27 V 18 6 EENG 2009 FS 2006 Part 5: Nodal Analysis = (45 – 27)2 / 6 = 54 W 164 Example 5 + Find va 1 mA 2 k va 1 k 3 k + – – 6V Solution: (Note that in this circuit the 6 V voltage source has one of its nodes connected to the reference node. This makes solution easier!) va is the only 1 mA unknown node voltage, so write KCL at node a: Node voltage already known! node A va + va 1 k – 2 k 3 k 6V + – 6V va/1000 + va/3000 + (va – 6)/2000 – 10–3 = 0 (6 + 2 + 3) va = 18 + 6 va = 2.18 V EENG 2009 FS 2006 Part 5: Nodal Analysis 165 Solution (cont.): va + 1 mA va i1k 1 k 2 k 3 k 6V + – – 6V In this circuit, the independent current source produces mA, the independent voltage source produces volts, and the resistors have values in k. Currents in the circuit will be in the mA range. For example, i1k = 2.18 V / 1 k = 2.18 mA. In these situations it is convenient when writing KCL to express the currents in mA and write the equation as follows: va / 1 + va/ 3 + (va – 6) / 2 – 1 = 0 EENG 2009 FS 2006 Part 5: Nodal Analysis 166 5.3 Supernodes A supernode is a set of nodes connected to each other by voltage sources, but not to the reference node by a path of voltage sources. Example 1 Identify the supernode. Solution: supernode 10 V + – v1 3A 13 5 v2 4 14 A Note that the supernode includes the component(s) in parallel with the voltage source. (It includes all of the circuit elements connected between the two nodes with the node voltages v1 and v2.) EENG 2009 FS 2006 Part 5: Nodal Analysis 167 Solution Procedure for Circuits with Supernodes Step 1. Write KCL equation for a surface enclosing the supernode. Step 2. Write a KVL equation relating the nodal voltages in the supernode (a constraint equation). Step 3. Solve the equations. Example 2. Find v1 and v2. Solution: 3A 10 V supernode + – v1 13 5 v2 4 14 A 1. KCL @ supernode: 3 + v1/5 + v2/4 – 14 = 0 2. Constraint equation: v1 – v2 = 10 3. Re-writing these two equations: 4 v1 + 5 v2 = 220 – v1 + v2 = – 10 Solving gives: v1 = 30 V, v2 = 20 V EENG 2009 FS 2006 Part 5: Nodal Analysis 168 Example 3 12 k Find v1 6V + – 6 k v1 3 k 14 mA 2 mA Solution: In this circuit there are two regular nodes and one supernode: supernode regular node v3 12 k v2 6 V v1 + – 6 k 14 mA 3 k 2 mA regular node The solution procedure in this case is: Step 1. Write the KCL equation for the supernode and the KCL equation for the regular node that was not chosen as the supernode. Step 2. Write the constraint equation. Step 3. Solve. EENG 2009 FS 2006 Part 5: Nodal Analysis 169 Solution (cont.) v3 6 k 12 k v2 6 V + – 14 mA v1 3 k 2 mA Step 1: KCL @ the supernode: (v2 – v3) / (12x103 ) + v1 / (3x103 ) + 2x10-3 = 0 Step 1 (cont.): KCL @ the non-reference node: v3 / (6x103 ) + (v3 – v2) / (12x103 ) – 14x10-3 = 0 Step 2: Constraint equation: v2 – v1 = 6 Step 3: Solving the 3 equations gives: v1 = 6 V v2 = 12 V v3 = 60 V EENG 2009 FS 2006 Part 5: Nodal Analysis 170 Example 4 Find vo 30 V 20 V + – + – 2 k 4 k + vo – 4 k + vo – 5 k Solution: vo 30 V 2 k + – 20 V + – v1 5 k v2 Step 1: KCL @ the supernode: v1 / 2x103 + v2 / 5x103 + vo / 4x103 = 0 Step 2: Constraint equations: vo – v1 = 30 vo – v2 = 20 Step 3: Solving the 3 equations gives: vo = 20 V v1 = – 10 V v2 = 0 V EENG 2009 FS 2006 Part 5: Nodal Analysis 171 5.4 Dependent Sources Dependent sources require constraint equations. Example 1. Find v1, v2, and v3. 2 7A 1 v1 + v1 – 1 v2 2 4 vx 4 1/2 v 3 – vx + 3 3/2 A Solution: KCL @ node 1: v1 / 2 + (v1–v2) / 1 – 4vx + 7 = 0 KCL @ node 2: (v2–v1) / 1 + v2 / 4 + (v2–v3) /(1/2) = 0 KCL @ node 3: (v3–v2) /(1/2) – 3/2 – 7 = 0 constraint equation: vx = v3 – v2 Four equations in four unknowns. The solution is: v1 = 24 V, v2 = 30.25 V, v2 = 26 V, vx = 4.25 V EENG 2009 FS 2006 Part 5: Nodal Analysis 172 Example 2. Find i1 and i2. 2 i1 v2 v1 15 A i1 2 3 i2 4 Solution: KCL @ node 1: v1 / 2 + 2 i1 + (v1–v2) / 3 = 15 KCL @ node 2: (v2–v1) / 3 + v2 / 4 – 2 i1 = 0 constraint: v1 / 2 = i1 Solving the 3 simultaneous equations in v1, v2, i1, and then using the relationship i2 = v2 / 4 gives: i1 = 7 A i2 = 8 A EENG 2009 FS 2006 Part 5: Nodal Analysis 173 1 Example 3. Find vx. – vx + 1/4 1/2 6A + – 3 vx 30 V 1/2 1/3 Solution: First, redraw to emphasize the nodes. Then choose a reference node and label the node voltages. Then write the KCL and constraint equations. v2 Note that for the chosen reference node there is no supernode present, but one of the node v1 voltages is now known to be –30 V. 1/4 1/2 1 6A + – 3 vx 1/2 EENG 2009 FS 2006 Part 5: Nodal Analysis – 30 V 30 V 1/3 174 Solution (cont.): v2 node 2 1/4 1/2 1 node 1 6A + – v1 3 vx 1/2 –30 V 30 V 1/3 node 1: – vx / 1 + v1 / (1/2) + 3 vx = 0 node 2: vx / 1 + v2 / (1/2) + (v2+30)/(1/4) – 6 = 0 constraint: v2 – v1 = vx Solving: v1 = –12 V v2 = 18 v vx = 6 v EENG 2009 FS 2006 Part 5: Nodal Analysis 175 Example 5. (Contains dependent sources and a supernode, too!) Find v and v1. 6A + 1 + v – v1 4 i1 6V – – + + – 1 2 1.5 v1 4 i1 Solution: First, identify the nodes and supernode. There is one supernode and +v two regular v nodes (including – the reference node). 6A vc node c + 1 6V + – 1 va 4 i1 – + 4 i1 2 v1 – vb supernode 1.5 v1 reference node EENG 2009 FS 2006 Part 5: Nodal Analysis 176 Solution (cont.): 6A vc 1 6V supernode v + v – + – va 1 Could also write –i1 here 4 i1 – + 4 i1 node c + 2 v1 v–b 1.5 v1 reference node KCL @ supernode: v / 1 + va / 4 – 1.5v1 + (vb –vc) / 1 – 6 = 0 KCL @ node c: (vc –vb) / 1 + vc / 2 + 6 = 0 Constraint equations: v – va = 6 vb – va = 4 i1 i1 = – v / 1 vc – vb = v1 Solving: v = – 2 V, v1 = – 4 V EENG 2009 FS 2006 Part 5: Nodal Analysis 6 unknowns: v v1 va vb vc i1 177 Example 6. 6i Find v 2 i – + 2 + v – 4A 8V + – 2 + – 4V Solution: Try working this out on your own. If you select the reference node carefully you can save yourself some effort. EENG 2009 FS 2006 Part 5: Nodal Analysis 178