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E ENG 2009

E ENG 2009. Part 5. Nodal Analysis 5.1 Nodes and Node Voltages 5.2 Nodal Analysis By Example 5.3 Supernodes 5.4 Dependent Sources.

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E ENG 2009

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  1. EENG 2009 • Part 5. Nodal Analysis • 5.1 Nodes and Node Voltages • 5.2 Nodal Analysis By Example • 5.3 Supernodes • 5.4 Dependent Sources EENG 2009 FS 2006 Part 5: Nodal Analysis

  2. 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). 5.1 Nodes and Node Voltages Nodes Example 1 Identify the nodes for the given circuit. 16  8  12  3 A 9 A 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: Solution: A single node! EENG 2009 FS 2006 Part 5: Nodal Analysis

  3. A node voltage associated with a given node is defined to be the voltage difference between the given node and a referencenode, 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. Identify a reference node and corresponding node voltages for the given circuit (whose nodes we found in the previous example). Node Voltages Example 2 16  8  12  3 A 9 A EENG 2009 FS 2006 Part 5: Nodal Analysis

  4. Using the results of Example 1, we draw the circuit with the nodes emphasized: Example 2 Solution: 16  8  12  3 A 9 A 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 node 2 8  12  3 A 9 A Reference node, ground, earth, “sea level” EENG 2009 FS 2006 Part 5: Nodal Analysis

  5. Measurement of Branch and Node Voltages voltmeter 16 V The voltage being measured is a branch voltage. + – v1 v2 16  1 2 12  8  9 A 3 A voltmeter voltmeter 64 V 48 V Node Voltages! + – + – 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

  6. Find the relationships among the branch voltages and the node voltages. Branch Voltages In Terms of Node Voltages Example 3 + v12 – v1 v2 2 1 + v13 – + v23 – 3 Solution: • There are three branch voltages, v13 ,v12 , and v23, and two node voltages, v1and v2 (not counting v3). • The branch voltages v13 and v23 are clearly equal to the node voltages v1and 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

  7. We can write the branch voltage v12 in terms of the node voltages v13 and v23 by applying KVL, as follows: Solution (cont.): + v12 – v1 v2 1 2 + v13 – + v23 – 3 • v12 = v13 – v23 = v1 – v2 • Note the correlation with the order of the subscripts: The subscripts of v12are 1 & 2 and the subscripts of the two node voltages being subtracted are 1 & 2. • In general, vj vk node j node k Memorize me! vjk = vj – vk EENG 2009 FS 2006 Part 5: Nodal Analysis

  8. Find the relationships among the branch current and the node voltages. Note that neither voltage-polarity markings nor a current reference arrow is shown on the diagram! Branch Currents In Terms of Node Voltages Example 4 R v1 v2 node 1 node 2 reference node Solution: • The branch current flowing from node 1to node 2 is: • i12 = ( v1 – v2 ) / R • The branch current flowing from node 2to node 1 is: • i21 = ( v2 – v1 ) / R • In general: R vj vk Note that to avoid confusion, ijk, ikj, and their corresponding reference directions are not shown on the diagram. node j node k ijk = ( vj – vk ) / R ikj = ( vk – vj ) / R EENG 2009 FS 2006 Part 5: Nodal Analysis

  9. 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

  10. Example 1 Find the values of the node voltages. R2 R1 is1 is2 R3 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

  11. Step 2. Write KCL at nodes 1 and 2. Solution (cont.): v1 v2 R2 node 1 node 2 R3 R1 is1 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

  12. Development of the AlgorithmicMethod for Writing Nodal- Analysis Equations Consider the previous circuit and its nodal equations: R2 v1 v2 is1 R1 R3 is2 • 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

  13. Now, using conductances: • (G1 + G2) v1 – G2 v2 = is1 (1) • – G2 v1 + (G2 + G3) v2 = – is2 (2) v1 R2 v2 is1 R1 R3 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

  14. Find v1 and v2. Example 2 16  + V1- + V2- 12  8  9 A 3 A Electrically the same as the node 2 v1 v2 Solution: 16  node 1 node 2 12  3 A 8  9 A • 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 = 9 • –1/16 v1 + (1/16 + 1/12) v2 = 3 • The solution is v1 = 64 V, v2 = 48 V. EENG 2009 FS 2006 Part 5: Nodal Analysis

  15. Note that these equations could have been written directly by using the algorithm! Solution (cont.) v1 v2 16  8  3A 12  9 A • (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 –1ENTER • a1,2 = – 16 –1ENTER • b1 = 9 ENTER • a2,1 = – 16 –1ENTER • a2,2 = 16 –1+ 12–1ENTER • b2 = 3 ENTER • SOLVE • x1 = 64.000 • x2 = 48.000 EENG 2009 FS 2006 Part 5: Nodal Analysis

  16. Find the voltage differences across the sources. Example 3. 5 A 4  4  6 A 2  3 A v2 Solution: node 2 5 A 4  v1 6 A 4  node 1 2  3 A • 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

  17. We still have to find the voltage differences across the three sources, as follows.* Example 3. (cont.) v2 = 20 V 5 A + v5A – 4  + v6A – v1 = – 4 V 6 A 4  + v3A – 2  3 A • 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

  18. 12  Find the current i and the power absorbed by the 6 resistor. Solution: Example 4. 6  i 4  5 A 6  18  v2 12  node 2 6  i node 1 4  v3 v1 5 A node 3 6  18  • 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

  19. To calculate i we apply Ohm’s Law (which requires calculating the branch voltage across the 4  resistor): • 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: Solution (cont.): i 4  v3 v1 12  v2 = 45 V p6 = (v2 – v3)2 / 6 = (45 – 27)2 / 6 = 54 W 6  4  5 A v1 v3 = 27 V 6  18  EENG 2009 FS 2006 Part 5: Nodal Analysis

  20. Find va Example 5 + va – 2 k 1 k 3 k 1 mA 6 V + – (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 unknown node voltage, so write KCL at node a: Solution: Node voltage already known! va node A + va – 2 k 1 k 3 k 6 V 1 mA 6 V + – • 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

  21. Solution (cont.): va + va – i1k 2 k 1 k 3 k 6 V 1 mA 6 V + – • 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

  22. 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. Identify the supernode. Example 1 Solution: supernode 10 V + – v1 v2 13  5  4  3 A 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

  23. 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 constraintequation). • Step 3. Solve the equations. • Find v1 and v2. Solution Procedure for Circuits with Supernodes Example 2. 10 V supernode + – Solution: v1 v2 13  5  4  14 A 3 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

  24. Find v1 Example 3 6 V v1 12 k + – 6 k 3 k 2 mA 14 mA In this circuit there are two regular nodes and one supernode: Solution: supernode 6 V v3 v2 v1 12 k regular node + – 6 k 3 k 2 mA 14 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

  25. Solution (cont.) 6 V v3 v2 v1 12 k + – 6 k 3 k 2 mA 14 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

  26. Example 4 Find vo 20 V 30 V + – + vo – + – 4 k 2 k 5 k Solution: vo + vo – 20 V 30 V + – + – 4 k v1 v2 2 k 5 k • 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

  27. 5.4 Dependent Sources Dependent sources require constraintequations. Find v1, v2, and v3. Example 1. 7 A v2 1  v1 1/2  v3 1 2 3 – vx + + v1 – 2  4  3/2 A 4 vx • 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 Solution: EENG 2009 FS 2006 Part 5: Nodal Analysis

  28. Find i1 and i2. Example 2. 2 i1 v2 v1 i2 i1 3  2  4  15 A 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

  29. Find vx. Example 3. 1  – vx + 1/4  1/2  6 A + – 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 voltages is now known to be –30 V. – vx + 1/4  1/2  1  6 A v1 – 30 V + – 3 vx 30 V 1/2  1/3  EENG 2009 FS 2006 Part 5: Nodal Analysis

  30. Solution (cont.): v2 node 2 – vx + 1/4  1/2 1 6 A node 1 v1 –30 V + – 3 vx 30 V 1/2  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

  31. Example 5. (Contains dependentsources and a supernode, too!) Find v and v1. 6 A + v1 – 1  2  4 i1 6 V – + + – + v – 1.5 v1 1  4  i1 First, identify the nodes and supernode. There is one supernode and two regular nodes (including the reference node). Solution: 6 A vc node c + v1 – 1  2  4 i1 6 V va vb v supernode – + + – + v – 1.5 v1 1  4  reference node i1 EENG 2009 FS 2006 Part 5: Nodal Analysis

  32. Solution (cont.): vc 6 A node c + v1 – 1  2  4 i1 6 V va vb v supernode – + + – + v – 1.5 v1 1  4  reference node Could also write –i1 here i1 • 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 • 6unknowns: • v • v1 • va • vb • vc • i1 EENG 2009 FS 2006 Part 5: Nodal Analysis

  33. Find v Example 6. 6 i i 2  – + 8 V 2  + – + v – 4 V 2  + – 4 A 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

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