EENG 2009 FS 2006 Part 5: Nodal Analysis 146 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 147 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). 9 A 16  8  12  3 A Identify the nodes for the given circuit. 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: 5.1 Nodes and Node Voltages Nodes Example 1 Solution: A single node!

EENG 2009 FS 2006 Part 5: Nodal Analysis 148 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. Identify a reference node and corresponding node voltages for the given circuit (whose nodes we found in the previous example). 9 A 16  8  12  3 A Node Voltages Example 2

EENG 2009 FS 2006 Part 5: Nodal Analysis 149 node 1 v1v1 v2v2 Reference node, ground, earth, “sea level” Using the results of Example 1, we draw the circuit with the nodes emphasized: 9 A 16  8  12  3 A Next choose the bottom node as the reference node, and designate the node voltages for the other two nodes as v 1 and v 2 : 9 A 16  8  12  3 A node 2 Example 2 Solution:

EENG 2009 FS 2006 Part 5: Nodal Analysis A 16  8  12  3 A 12 v1v1 v2v2 64 V48 V 16 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 v 1 and v 2. voltmeter + – The voltage being measured is a branch voltage. Node Voltages! Measurement of Branch and Node Voltages

EENG 2009 FS 2006 Part 5: Nodal Analysis v1v1 v2v2 3 + v 12 – + v 23 – + v 13 – Find the relationships among the branch voltages and the node voltages. There are three branch voltages, v 13, v 12, and v 23, and two node voltages, v 1 and v 2 (not counting v 3 ). The branch voltages v 13 and v 23 are clearly equal to the node voltages v 1 and v 2 : v 13 = v 1 v 23 = v 2 The branch voltage v 12 is a combination of the node voltages v 1 and v 2. Branch Voltages In Terms of Node Voltages Example 3 Solution:

EENG 2009 FS 2006 Part 5: Nodal Analysis 152 v 12 = v 13 – v 23 = v 1 – v 2 Note the correlation with the order of the subscripts: The subscripts of v 12 are 1 & 2 and the subscripts of the two node voltages being subtracted are 1 & 2. In general, We can write the branch voltage v 12 in terms of the node voltages v 13 and v 23 by applying KVL, as follows: 12 v1v1 v2v2 3 + v 12 – + v 23 – + v 13 – node jnode k vjvj vkvk Memorize me! Solution (cont.): v jk = v j – v k

EENG 2009 FS 2006 Part 5: Nodal Analysis 153 R v1v1 v2v2 The branch current flowing from node 1 to node 2 is: i 12 = ( v 1 – v 2 ) / R The branch current flowing from node 2 to node 1 is: i 21 = ( v 2 – v 1 ) / R In general: node 1node 2 reference node 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 Solution: node jnode k vjvj vkvk i jk = ( v j – v k ) / R i kj = ( v k – v j ) / R R Note that to avoid confusion, i jk, i kj, and their corresponding reference directions are not shown on the diagram.

EENG 2009 FS 2006 Part 5: Nodal Analysis 154 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. 5.2 Nodal Analysis By Example

EENG 2009 FS 2006 Part 5: Nodal Analysis 155 i s1 R1R1 R3R3 i s2 v1v1 v2v2 R2R2 Find the values of the node voltages. Step 1. There are 3 nodes. Choose the bottom node as the reference node, and designate the other two nodes as v 1 and v 2. i s1 R1R1 R3R3 i s2 With 3 nodes and one of them as the reference node, there are two nodal equations needed. R2R2 node 1 node 2 node 3 Example 1 Solution:

EENG 2009 FS 2006 Part 5: Nodal Analysis 156 v1v1 v2v2 i s1 R1R1 R3R3 i s2 Step 2. Write KCL at nodes 1 and 2. KCL at node 1 (Summing the currents flowing out of node 1): v 1 / R 1 + (v 1 – v 2 ) / R 2 – i s1 = 0 At node 2: (v 2 – v 1 ) / R 2 + v 2 / R 3 + i s2 = 0 These are the equations to solve for v 1 and v 2. We can put them in orderly form (and discover a shortcut to writing them), as we show next. R2R2 node 1node 2 Solution (cont.):

EENG 2009 FS 2006 Part 5: Nodal Analysis 157 Consider the previous circuit and its nodal equations: v 1 / R 1 + (v 1 – v 2 ) / R 2 – i s1 = 0 (v 2 – v 1 ) / R 2 + v 2 / R 3 + i s2 = 0 Collect terms in v 1 and v 2 and put the independent-source terms on the RHS: (1 / R / R 2 ) v 1 – 1/ R 2 v 2 = i s1 – 1/ R 2 v 1 + (1 / R / R 3 ) v 2 = – i s2 v1v1 v2v2 i s1 R1R1 R3R3 i s2 R2R2 Development of the Algorithmic Method for Writing Nodal- Analysis Equations

EENG 2009 FS 2006 Part 5: Nodal Analysis 158 Now, using conductances: (G 1 + G 2 ) v 1 – G 2 v 2 = i s1 (1) – G 2 v 1 + (G 2 + G 3 ) v 2 = – i s2 (2) Observe that the following algorithm is valid: In (1), the coefficient of v 1 =  (conductances connected to node 1) In (1), the coefficient of v 2 = –  (conductances between nodes 1 & 2) In (2), the coefficient of v 2 =  (conductances connected to node 2) In (2), the coefficient of v 1 = –  (conductances between nodes 1 & 2) In (1) and (2) the RHS =  (current sources entering the node) v1v1 v2v2 i s1 R1R1 R3R3 i s2 R2R2

EENG 2009 FS 2006 Part 5: Nodal Analysis A +V1+V1 3 A Find v 1 and v 2. 8  16  12  Solution: 9 A 3 A 8  12  v1v1 v2v2 node 1: v 1 / 8 + (v 1 – v 2 ) / 16 – 9 = 0 node 2: (v 2 – v 1 ) / 16 + v 2 /12 – 3 = 0 Collecting terms: (1/8 + 1/16) v 1 – 1/16 v 2 = 9 –1/16 v 1 + (1/16 + 1/12) v 2 = 3 The solution is v 1 = 64 V, v 2 = 48 V. 16  Electrically the same as the node 2 node 1 node 2 Example 2 +V2+V2

EENG 2009 FS 2006 Part 5: Nodal Analysis 160 Note that these equations could have been written directly by using the algorithm! 9 A 3 A3 A 8  12  v1v1 v2v2 (1/8 + 1/16)v 1 - 1/16 v 2 = 9 –1/16 v 1 + (1/16 + 1/12) v 2 = 3 The TI-85 & TI-86 keystrokes for solving this equation are : 2nd SIMULT Number = 2 ENTER a1,1 = 8 – –1 ENTER a1,2 = – 16 –1 ENTER b1 = 9 ENTER a2,1 = – 16 –1 ENTER a2,2 = 16 – –1 ENTER b2 = 3 ENTER SOLVE x 1 = x 2 =  Solution (cont.)

EENG 2009 FS 2006 Part 5: Nodal Analysis 161 Find the voltage differences across the sources. 3 A 4  5 A 2  4  Solution: 3 A 4  5 A 2  v2v2 v1v1 nodes 1 and 2: (v 1 – v 2 ) / v 1 / 2 – 3 = 0 (v 2 – v 1 ) / 4 – 5 – 6 + v 2 / 4 = 0 Solution: v 1 = – 4 V, v 2 = 20 V 4  6 A node 1 node 2 Example 3.

EENG 2009 FS 2006 Part 5: Nodal Analysis 162 We still have to find the voltage differences across the three sources, as follows.* 3 A 4  5 A 2  v 2 = 20 V v 1 = – 4 V 4  6 A + v 3A – + v 5A – + v 6A – v 3A = v 1 = – 4 V v 6A = v 2 = 20 V v 5A = v 2 – v 1 = 24 V Example 3. (cont.) * 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 163 Find the current i and the power absorbed by the 6  resistor. Solution: 6  12  5 A 18  4  i 6  12  5 A 18  4  i v2v2 v3v3 v1v1 1 : (v 1 – v 3 )/4 + (v 1 – v 2 )/12 + v 1 /6 = 0 2 : (v 2 – v 1 )/12 + (v 2 – v 3 )/6 – 5 = 0 3 : v 3 /18 + (v 3 – v 1 )/4 + (v 3 – v 2 )/6 = 0 Solving for the node voltages gives: v 1 = 21 V, v 2 = 45 V, v 3 = 27 V 6  node 1 node 2 node 3 Example 4.

EENG 2009 FS 2006 Part 5: Nodal Analysis 164 To calculate i we apply Ohm’s Law (which requires calculating the branch voltage across the 4  resistor): i = ( v 3 – v 1 ) / 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: v3v3 v1v1 4  i 6  12  5 A 18  4  v 2 = 45 V v 3 = 27 V v1v1 p 6 = (v 2 – v 3 ) 2 / 6 = (45 – 27) 2 / 6 = 54 W 6  Solution (cont.):

EENG 2009 FS 2006 Part 5: Nodal Analysis 165 Find v a 6 V 3 k  1 mA 2 k  +–+– 1 k  +va–+va– (Note that in this circuit the 6 V voltage source has one of its nodes connected to the reference node. This makes solution easier!) v a is the only unknown node voltage, so write KCL at node a: 6 V 3 k  1 mA 2 k  +–+– 1 k  +va–+va– 6 V vava v a / v a / (v a – 6)/2000 – 10 –3 = 0 ( ) v a = v a = 2.18 V node A Example 5 Solution: Node voltage already known!

EENG 2009 FS 2006 Part 5: Nodal Analysis V 3 k  1 mA 2 k  +–+– 1 k  +va–+va– 6 V vava 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, i 1k  = 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: v a / 1 + v a / 3 + (v a – 6) / 2 – 1 = 0 Solution (cont.): i 1k 

EENG 2009 FS 2006 Part 5: Nodal Analysis A 5  14 A 4  13  + – 10 V supernode 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 v 1 and v 2.) v1v1 v2v2 5.3 Supernodes Example 1 Solution: 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.

EENG 2009 FS 2006 Part 5: Nodal Analysis 168 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. Find v 1 and v 2. 3 A 5  14 A 4  13  + – 10 V supernode 1. supernode: 3 + v 1 /5 + v 2 /4 – 14 = 0 2. Constraint equation: v 1 – v 2 = Re-writing these two equations: 4 v v 2 = 220 – v 1 + v 2 = – 10 Solving gives: v 1 = 30 V, v 2 = 20 V v1v1 v2v2 Solution Procedure for Circuits with Supernodes Example 2. Solution:

EENG 2009 FS 2006 Part 5: Nodal Analysis 169 Find v 1 14 mA 6 k  2 mA 3 k  12 k  + – 6 V v1v1 In this circuit there are two regular nodes and one supernode: 14 mA 6 k  2 mA 3 k  12 k  + – 6 V v1v1 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. v3v3 v2v2 supernode regular node Example 3 Solution:

EENG 2009 FS 2006 Part 5: Nodal Analysis 170 Solution (cont.) Step 1: the supernode: (v 2 – v 3 ) / (12 x 10 3 ) + v 1 / (3 x10 3 ) + 2 x = 0 Step 1 (cont.): the non-reference node: v 3 / (6 x 10 3 ) + (v 3 – v 2 ) / (12 x 10 3 ) – 14 x = 0 Step 2: Constraint equation: v 2 – v 1 = 6 Step 3: Solving the 3 equations gives: v 1 = 6 V v 2 = 12 V v 3 = 60 V 14 mA 2 mA 3 k  12 k  6 V v1v1 v3v3 v2v2 6 k  + –

EENG 2009 FS 2006 Part 5: Nodal Analysis 171 Example 4 2 k  4 k  30 V +–+– 5 k  20 V +–+– +vo–+vo– Find v o Solution: 2 k  4 k  30 V +–+– 5 k  20 V +–+– +vo–+vo– vovo v1v1 v2v2 Step 1: the supernode: v 1 / 2 x v 2 / 5 x v o / 4 x 10 3 = 0 Step 2: Constraint equations: v o – v 1 = 30 v o – v 2 = 20 Step 3: Solving the 3 equations gives: v o = 20 V v 1 = – 10 V v 2 = 0 V

EENG 2009 FS 2006 Part 5: Nodal Analysis Dependent Sources Dependent sources require constraint equations. Find v 1, v 2, and v 3. 4 v x 3/2 A 4  1  7 A v3v3 v1v1 v2v2 2  1/2  – v x v1–+v1– node 1: v 1 / 2 + (v 1 –v 2 ) / 1 – 4v x + 7 = 0 node 2: (v 2 –v 1 ) / 1 + v 2 / 4 + (v 2 –v 3 ) /( 1/2) = 0 node 3: (v 3 –v 2 ) /( 1/2) – 3/2 – 7 = 0 constraint equation: v x = v 3 – v 2 Four equations in four unknowns. The solution is: v 1 = 24 V, v 2 = V, v 2 = 26 V, v x = 4.25 V Example 1. Solution:

EENG 2009 FS 2006 Part 5: Nodal Analysis A 3  2 i 1 v1v1 v2v2 2  4  i2i2 i1i1 Find i 1 and i 2. node 1: v 1 / i 1 + (v 1 –v 2 ) / 3 = 15 node 2: (v 2 –v 1 ) / 3 + v 2 / 4 – 2 i 1 = 0 constraint: v 1 / 2 = i 1 Solving the 3 simultaneous equations in v 1, v 2, i 1, and then using the relationship i 2 = v 2 / 4 gives: i 1 = 7 A i 2 = 8 A Example 2. Solution:

EENG 2009 FS 2006 Part 5: Nodal Analysis A 1  1/2  – v x + 1/2  1/3  3 v x + – 30 V 1/4  Find v x. First, redraw to emphasize the nodes. Then choose a reference node and label the node voltages. Then write the KCL and constraint equations. + – 30 V 1/2  1/3  3 v x v1v1 – 30 V v2v2 – v x + 1  6 A 1/4  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. Example 3. Solution:

EENG 2009 FS 2006 Part 5: Nodal Analysis – 30 V 1/2  1/3  3 v x v1v1 –30 V v2v2 – v x + 11 6 A 1/4  Solution (cont.): node 1: – v x / 1 + v 1 / (1/2) + 3 v x = 0 node 2: v x / 1 + v 2 / (1/2) + (v 2 +30)/( 1/4 ) – 6 = 0 constraint: v 2 – v 1 = v x Solving: v 1 = –12 V v 2 = 18 v v x = 6 v node 2 node 1

EENG 2009 FS 2006 Part 5: Nodal Analysis 176 (Contains dependent sources and a supernode, too!) Find v and v 1. 6 A 6 V + – 4 i 1 1  +v–+v– – + 4  1.5 v 1 i1i1 +v1–+v1– 2  6 A 6 V + – 4 i 1 1  +v–+v– – + 4  1.5 v 1 i1i1 +v1–+v1– 2  First, identify the nodes and supernode. There is one supernode and two regular nodes (including the reference node). v vava vbvb vcvc supernode node c reference node Example 5. Solution:

EENG 2009 FS 2006 Part 5: Nodal Analysis 177 Solution (cont.): supernode: v / 1 + v a / 4 – 1.5v 1 + (v b –v c ) / 1 – 6 = 0 node c: (v c –v b ) / 1 + v c / = 0 Constraint equations: v – v a = 6 v b – v a = 4 i 1 i 1 = – v / 1 v c – v b = v 1 Solving: v = – 2 V, v 1 = – 4 V 6 unknowns: v v 1 v a v b v c i 1 Could also write –i 1 here 6 A 6 V + – 4 i 1 1  +v–+v– – + 4  1.5 v 1 i1i1 +v1–+v1– 2  v vava vbvb vcvc supernode node c reference node

EENG 2009 FS 2006 Part 5: Nodal Analysis 178 Find v Example 6. 8 V + – 6 i – + 2  i 4 V 4 A 2  +–+– + v – Try working this out on your own. If you select the reference node carefully you can save yourself some effort. Solution: