07-Nodal Analysis Text: ECEGR 210 Electric Circuits I
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1 07Nodal Analysis Text: ECEGR 210 Electric Circuits I
2 Overview Introduction Nodal Analysis Nodal Analysis with Voltage Sources Dr. Louie 2
3 Basic Circuit Laws Ohm s Law Introduction Kirchhoff s Voltage Law (KVL) Kirchhoff s Current Law (KCL) Now we seek to analyze any linear circuit systematically using Nodal analysis Mesh analysis Dr. Louie 3
4 Introduction Express circuit as a system of linear equations Solve linear equations Substitution Gaussian elimination Cramer s Rule Numerical methods Others y11 y12 y13 x1 b1 y21 y22 y23 x2 b2 y 31 y32 y33 x3 b3 Remember: for a unique solution there must be an equal number of independent equations as unknowns Dr. Louie 4
5 Introduction Let X be a vector of variables to be solved for Nodal Analysis: X: node voltages Y: conductances b: currents Mesh Analysis: X: loop currents Y: resistances b: voltages y11 y12 y13 x1 b1 y21 y22 y23 x2 b2 y 31 y32 y33 x3 b3 Dr. Louie 5
6 Nodal Analysis Steps to solve a circuit with N nodes 1. Assign a reference node 2. Assign a variable to the voltages at all nodes wrt the reference node (N1 variables) 3. Apply KCL to each of the N1 nonreference nodes to generate N1 equations 4. Use Ohm s Law to express currents as functions of node voltages 5. Solve resulting simultaneous equations Dr. Louie 6
7 Nodal Analysis Let I 1, I 2, R 1, R 2, R 3 be known (given) Solve for the voltages at each node I 2 R 2 I 1 R 1 R 3 Dr. Louie 7
8 Step 1: Assign Reference Node Recall: all voltage is a relative measure Reference node is also known as ground In power systems it is actually the ground (earth) In electronics it could be the metallic chassis Or arbitrary I 2 R 2 I 1 R 1 R 3 (usually the bottom of circuit) Dr. Louie 8
9 Step 2: Assign Variables to Nodes There are two remaining nodes (variables) Assign voltages and polarities wrt reference node No need to write voltage across R 2 Does not add an independent equation (no new information) Can solve later as V 1 V 2 I 2 + I 1 V 1 x R 2 R 1 R 3 x + V 2 Dr. Louie 9
10 Step 3: Apply KCL KCL gives one equation per node (two equations) I 1 = I 2 + i 1 + i 2 I 2 = i 3 i 2 I 2 i 2 I 1 i 1 i 3 Dr. Louie 10
11 Step 4: Apply Ohm s Law Need to write i 1, i 2, i 3 in terms of the node voltages Remember sign convention i 1 = (V 1 0)/R 1 i 2 = (V 1 V 2 )/R 2 I 2 i 3 = (V 2 0)/R 3 V 1 i 2 x x V 2 I 1 i 1 i 3 Dr. Louie 11
12 Recap Step 5: Solve Equations two variables (V 1, V 2 ) two KCL equations KCL equations use intermediate variables i 1, i 2, i 3 I 1 = I 2 + i 1 + i 2 I 2 = i 3 i 2 Use substitution to express KCL in terms of voltages i 1 = (V 1 0)/R 1 i 2 = (V 1 V 2 )/R 2 i 3 = (V 2 0)/R 3 Dr. Louie 12
13 Via substitution: Step 5: Solve Equations I 1 = I 2 + V 1 /R 1 + (V 1 V 2 )/R 2 I 2 = V 2 /R 3 (V 1 V 2 )/R 2 Can we solve this system of equations? Yes! Two independent equations, two unknowns The rest is just math Dr. Louie 13
14 In matrix form Step 5: Solve Equations I 1 = I 2 + V 1 /R 1 + (V 1 V 2 )/R 2 I 2 = V 2 /R 3 (V 1 V 2 )/R R 1 R2 R2 V 1 I1 I V2 I2 R2 R3 R2 Many methods of solving linear equations Dr. Louie 14
15 Example Let I 1 = 10A I 2 = 5A R 1 = 6Ω R 2 = 4Ω R 3 = 2Ω Find all voltages I 2 i 2 I 1 i 1 i 3 Dr. Louie 15
16 Example Equations: I 1 = I 2 + V 1 /R 1 + (V 1 V 2 )/R 2 I 2 = V 2 /R 3 (V 1 V 2 )/R 2 Using circuit values: 10 = 5 + V 1 /6 + (V 1 V 2 )/4 5 = V 2 /2 (V 1 V 2 )/4 Dr. Louie 16
17 Example Starting with the second equation 5x4 = [V 2 /2 (V 1 V 2 )/4]x4 Yields 20 = 2V 2 (V 1 V 2 ) = 3V 2 V 1 Now the first equation: 10x12 = [5 + V 1 /6 + (V 1 V 2 )/4]x12 120=60 + 2V 1 + 3V 1 3V 2 60 = 5V 1 3V 2 Dr. Louie 17
18 Example 20 = V 1 + 3V 2 60 = 5V 1 3V 2 Solve using elimination 20x5 = [V 1 + 3V 2 ]x5 = 100 = 5V V 2 60 = 5V 1 3V = 0 +12V 2 Yields: V 2 = V V 1 = 20V Dr. Louie 18
19 Solving in matrix form: Example R 1 R2 R2 V 1 I1 I V2 I2 R2 R3 R V V Can use Matlab, Gaussian elimination Cramer s Rule, matrix inversion Dr. Louie 19
20 Solution by Matrix Inversion V V G (matrix) V1 5 G V 2 5 V V G where G For small matrices use inv() command in Matlab Dr. Louie 20
21 Nodal Analysis with Voltage Sources Nodal analysis requires knowledge of current through elements Recall that the current through a voltage source is not readily known Example: Current through resistor is I = (V a V b )/R But what is the current through a voltage source? How is the R in the equation to be interpreted? + V + Dr. Louie 21
22 Nodal Analysis with Voltage Sources If the voltage source is connected to the reference node, then it can be easily included in nodal analysis I 2 i 2 I 1 i 1 + V 3 Dr. Louie 22
23 Nodal Analysis with Voltage Sources Only one node with unknown voltage I 1 = I 2 + V 1 /R 1 + (V 1 V 3 )/R 2 Compare this to earlier example (2 eqns, 2 unknowns) I 2 R 2 I 1 + V 1 x R 1 + V 3 Dr. Louie 23
24 Nodal Analysis with Voltage Sources If voltage source not connected to reference: Move the reference so it is! (only works if there is one voltage source) Or Use Supernode concept Dr. Louie 24
25 + + Nodal Analysis with Voltage Sources Supernode: a closed surface containing a voltage source and its two nodes AND any elements in parallel with it I I node I I Supernodes Current into supernode = current out of supernode Also applies to dependent voltage sources Dr. Louie 25
26 Nodal Analysis with Voltage Sources Properties of a supernode Voltage source inside the supernode provides a constraint equation needed to solve for node voltages A supernode has no voltage of its own There is only one KCL equation for the supernode, but two unknown node voltages Additional equation is found from application of KVL Dr. Louie 26
27 + Nodal Analysis with Voltage Sources Find the node voltages 10W 2V 2A 2W 4W 7A assume reference is here Dr. Louie 27
28 + Nodal Analysis with Voltage Sources Make a supernode out of the voltage source Write KCL for supernode 2 = i 1 + i (only one equation since it is a supernode) 10W 2V 2A 2W 4W i 1 i 2 7A assume reference is here Dr. Louie 28
29 + Nodal Analysis with Voltage Sources Write i 1, i 2 in terms of node voltage i 1 = (v 1 0)/2 i 2 = (v 2 0)/4 10W 2V v 1 v 2 2A 2W 4W i 1 i 2 7A assume reference is here Dr. Louie 29
30 + Nodal Analysis with Voltage Sources Substitute into supernode KCL 2 = i 1 + i (supernode KCL) 2 = 0.5v v (after substitution) 8 = 2v 1 + v (multiplying by 4) 10W 2V v 1 v 2 2A 2W i 1 4W i 2 7A Dr. Louie 30
31 + Nodal Analysis with Voltage Sources Apply KVL around loop v v 2 = 0 (second equation) 10W 2V v 1 v 2 2A 2W i 1 4W i 2 7A Dr. Louie 31
32 + Nodal Analysis with Voltage Sources How many independent equations? 8 = 2v 1 + v v v 2 = 0 How many unknowns? v 1, v 2 Solve! 10W 2V v 1 v 2 2A 2W i 1 4W i 2 7A Dr. Louie 32
33 + Nodal Analysis with Voltage Sources Solving 8 = 2v 1 + v v 1 = 5.333V v 2 = 7.333V 10W 2V v 1 v 2 2A 2W i 1 4W i 2 7A Dr. Louie 33
34 + Example Find the node voltages in the circuit shown 10V 2W 2W 4W 5A 8W Dr. Louie 34
35 + Example 3 unknown voltages: V 1, V 2, V 3 Need three independent equations (that do not introduce new variables) 10V V 1 2W V 2 2W V 3 4W 5A 8W Dr. Louie 39
36 + Identify the supernode Example 10V V 1 2W V 2 2W V 3 4W 5A 8W Dr. Louie 40
37 + Example Write KCL for the supernode i 5 = i 1 +i 4 +i 3 (supernode) Write KCL for the other node 5+i 4 = i 5 (node 2) 10V 4W V 1 i 4 2W 2W i 5 V 2 i 1 5A i 3 V 3 8W Dr. Louie 41
38 + Example Express currents in terms of voltages i 1 = (V 1 0)/4 i 3 = (V 3 0)/8 i 4 = (V 1 V 2 )/2 i 5 = (V 2 V 3 )/2 4W V 1 10V i 4 2W 2W i 5 V 2 i 1 5A i 3 V 3 8W Dr. Louie 42
39 + Have two equations Example KCL for node 2, supernode Need one more independent equation Do KVL around top loop 10 = 2i 4 + 2i 5 4W V 1 10V i 4 2W 2W i 5 V 2 i 1 5A i 3 V 3 8W Dr. Louie 43
40 5 + i 4 = i 5 (node 2) i 6 = i 1 + i 5 + i 3 (supernode) Example i 1 = (V 1 0)/4 = 0.25V 1 i 3 = (V 3 0)/8 = 0.125V 3 i 4 = (V 1 V 2 )/2 = 0.5V 1 0.5V 2 i 5 = (V 2 V 3 )/2 = 0.5V 2 0.5V V 1 0.5V 2 = 0.5V 2 0.5V 3 (solving node 2 eqn) 10 + V 1 V 2 = V 2 V 3 0.5V 2 0.5V 3 = 0.25V V 1 0.5V V 3 (supernode eqn) 4V 2 4V 3 = 2V 4 + 4V 1 4V 2 + V 3 10 = 2i 4 + 2i 5 (KVL of top loop) 10 = V 1 V 2 + V 2 V 3 Dr. Louie 44
41 Example 10 + V 1 V 2 = V 2 + V 3 (node 2) 10 = V 1 + 2V 2 + V 3 (after rearranging) 4V 2 4V 3 = 2V 1 + 4V 1 4V 2 V 3 (supernode) 0 = 6V 1 8V 2 + 3V 3 (after rearranging) 10 = V 1 V 2 + V 2 V 3 (KVL of top loop) 10 = V 1 + 0V 2 V 3 (after rearranging) Solving V 1 = 12.22V V 2 = 10V V 3 = 2.22V V V V 3 10 Dr. Louie 45
42 Example Find V x + 2A 10W V x 20W 0.2V x Dr. Louie 46
43 Example One node, write equation from KCL 2 =I 1 + I V x From Ohm s Law I 1 = V x /10 I 2 = V x /20 Solving: 2 = 0.1V x V x + 0.2V x V x = 5.71V 2A 10W I 1 + V x 20W I 2 0.2V x Dr. Louie 47
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