EE201 Circuit Theory Ch 3. Network Theorems (Part 1) Bongjin Kim Associate Professor Electrical Engineering, KAIST *Lecture materials are modified version of the contents provided by Prof. Minkyu Je (KAIST) Outline ❑ Nodal Analysis – Circuits Containing Only Independent Current Sources – Circuits Containing Dependent Current Sources – Circuits Containing Independent Voltage Sources – Circuits Containing Dependent Voltage Sources ❑ Loop Analysis – Circuits Containing Only Independent Voltage Sources – Circuits Containing Independent Current Sources – Circuits Containing Dependent Sources 2 Nodal Analysis ❑ In a nodal analysis, the variables in the circuit are selected to be the node voltages. – The node voltages are defined with respect to a common point in the circuit. – One node is selected as the reference node, and all other node voltages are defined with respect to that node. – Quite often, this node is the one to which the largest number of branches are connected. – It is commonly called ground because it is said to be at ground-zero potential, and it sometimes represents the chassis or ground line in a practical circuit. 3 Nodal Analysis ❑ Value of knowing all the node voltages in a network The bottom node is selected as the reference and labeled with the ground symbol. VS, Va, Vb and Vc are all measured with respect to the bottom node. Once these node voltages are known, we know the voltage across every element in the network. Thus, we can immediately calculate any branch current or the power supplied or absorbed by any element. Nodal and Loop Analysis Techniques 4 Nodal Analysis ❑ Value of knowing all the node voltages in a network V1 = VS − Va = 12 − 3 = 9 V 3 3 = V 2 2 3 3 9 V5 = Vb − Vc = − = V 2 8 8 V3 = Va − Vb = 3 − I1 = VS − Va 9 = = 1 mA 9k 9k I3 = Va − Vb 3 2 1 = = mA 3k 3k 2 I5 = Vb − Vc 9 8 1 = = mA 9k 9k 8 I2 = Va − 0 3 1 = = mA 6k 6k 2 I4 = Vb − 0 3 2 3 = = mA 4k 4k 8 Nodal and Loop Analysis Techniques 5 Nodal Analysis ❑ As a general rule, if we know the node voltages in the circuit, we can calculate the current through any resistive element using Ohm’s law: I= (Vm − Vn ) . R – Note carefully that both voltages Vm and Vn are measured with respect to the same point, that is, ground. – In a nodal analysis, this concept is central to the manner in which we write the equations necessary to determine all the node voltages. Nodal and Loop Analysis Techniques 6 Nodal Analysis ❑ In a nodal analysis, we employ KCL equations in such a way that the variables contained in these equations are the unknown node voltages of the network. – One of the nodes in an N-node circuit is selected as the reference node, and the voltages at all the remaining N − 1 nonreferernce nodes are measured with respect to this reference node. – Exactly N − 1 linearly independent KCL equations are required to determine the N − 1 unknown node voltages. – Therefore, once one of the nodes in an N-node circuit has been selected as the reference node, our task is reduced to identifying the remaining N − 1 nonreference nodes and writing one KCL equation at each of them. Nodal and Loop Analysis Techniques 7 Circuits w/ Indep. Current Sources ❑ Consider the network shown below. – This network contains three nodes, and thus N − 1 = 3 − 1 = 2 linearly independent KCL equations are required to determine the N − 1 = 2 unknown node voltages. – We select the bottom node as the reference node, and then the voltage at the two remaining nodes labeled v1 and v2 are measured with respect to this node. Nodal and Loop Analysis Techniques 8 Circuits w/ Indep. Current Sources ❑ Consider the network shown below. – Applying KCL at node 1 yields − i A + i 1 + i 2 = 0. Applying KCL at node 2 yields − i2 + iB + i3 = 0, or Using Ohm’s law, − G2 (v 1 − v 2 ) + iB + G3 (v 2 − 0) = 0 − i A + G1(v 1 − 0) + G2 (v 1 − v 2 ) = 0, or which can be expressed as (G1 + G2 )v 1 − G2v 2 = i A Nodal and Loop Analysis Techniques − G2v 1 + (G2 + G3 )v 2 = − iB . 9 Circuits w/ Indep. Current Sources ❑ Consider the network shown below. – Therefore, the following two equations need to be solved for the two unknown node voltages v1 and v2. (G1 + G2 )v 1 − G2v 2 = i A − G2v 1 + (G2 + G3 )v 2 = − iB – Three techniques will be demonstrated for solving linearly independent simultaneous equations: Gaussian elimination, matrix analysis, and MATLAB mathematical software package. Nodal and Loop Analysis Techniques 10 Example ❑ Let us determine all node voltages and branch currents. 6 k 4 mA 12 k 6 k 1 mA Using the parameter values, the equations become 1 1 1 V1 + − V2 = 1m and 12k 6k 6k 1 1 1 − V1 + V2 + = −4m. 6k 6k 6k The equations can be written as V1 V2 − = 1m and 4k 6k Nodal and Loop Analysis Techniques − V1 V2 + = −4m. 6k 3k 11 Example ❑ Let’s use Gaussian elimination to solve V1 V2 V V − = 1m and − 1 + 2 = −4m. 4k 6k 6k 3k We solve the first equation for V1 in terms of V2: 2 V1 = V2 + 4. 3 This value is then substituted into the second equation to yield − 1 2 V V2 + 4 + 2 = −4m or V2 = −15 V. 6k 3 3k This value for V2 is substituted back into the equation for V1 in terms of V2: 2 V1 = V2 + 4 = −6 V. 3 Nodal and Loop Analysis Techniques 12 Example ❑ Let’s use matrix analysis to solve V1 V2 V V − = 1m and − 1 + 2 = −4m. 4k 6k 6k 3k Using matrix analysis, we can write GV = I, where in this case 1 G = 4k 1 − 6k − 1 6k , V = V1 , and I = 1m . V − 4m 1 2 3k The solution to the matrix equation is V = G−1I and therefore, 1 V1 4k V = 1 2 − 6k −1 1 − 1m 6k . 1 − 4m 3k To calculate the inverse of G, we need the adjoint and the determinant. Nodal and Loop Analysis Techniques 13 Example ❑ Let’s use matrix analysis to solve V1 V2 V V − = 1m and − 1 + 2 = −4m. 4k 6k 6k 3k The adjoint is 1 Adj G = 3k 1 6k 1 6k , 1 4k and the determinant is 1 1 1 1 1 |G | = − − − = . 2 4k 3k 6k 6k 18k Therefore, 1 V1 2 3k = 18 k 1 V 2 6k Nodal and Loop Analysis Techniques 4 1 1 − 6k 1m = 18k 2 3k 2 6k 2 = − 6 . 1 1 − 4m 1 − 15 − 2 2 4k 6k k 14 Example ❑ Let’s use MATLAB to solve V1 V2 V V − = 1m and − 1 + 2 = −4m. 4k 6k 6k 3k Multiplying both equations by 12k yields 3V1 − 2V2 = 12 and − 2V1 + 4V2 = −48. In matrix form, the equation is 3 − 2 V1 12 − 2 V = − 48 . 4 2 Then the data entries and solution using MATLAB are as follows: >> G = [ 3 G = 3 -2 -2; -2 -2 4 Nodal and Loop Analysis Techniques 4 ] >> I = [ 12; I = 12 -48 -48 ] >> V = inv(G)*I V = -6.0000 -15.0000 15 Example ❑ Knowing the node voltages, we can determine all the currents using Ohm’s law. 6 k 12 k 1 mA I1 = V1 − 6 1 = = − mA R1 12k 2 I2 = V1 − V2 − 6 − (−15) 3 = = mA R2 6k 2 I3 = V2 − 15 5 = = − mA R3 6k 2 Nodal and Loop Analysis Techniques 4 mA 6 k 16 Circuits w/ Indep. Current Sources ❑ Let’s examine the network shown below. – This network has four nodes. – The node at the bottom of the circuit is selected as the reference node and labeled with the ground symbol. – Since N = 4, N − 1 = 3 linearly independent KCL equations will be required to determine the three unknown nonreference node voltages labeled v1, v2, and v3. Nodal and Loop Analysis Techniques 17 Circuits w/ Indep. Current Sources ❑ Let’s examine the network shown below. Node 2: − i 2 + i 4 − i 5 = 0. v 2 − v1 v 2 − 0 v 2 − v 3 + + =0 R2 R4 R5 1 1 1 1 1 − v 3 + v 2 + + =0 R2 R R R R 2 4 5 5 – Node 1: − v1 – Node 1: Node 3: i 1 − i A + i 2 − i 3 = 0. i 3 + i 5 + i B = 0. v1 − 0 v − v 2 v1 − v 3 − iA + 1 + =0 R1 R2 R3 v 3 − v1 v 3 − v 2 + + iB = 0 R3 R5 1 1 1 1 1 − v 2 v 1 + + − v3 = iA R R R R R 1 2 3 2 3 − v1 Nodal and Loop Analysis Techniques 1 1 1 1 = − iB − v2 + v 3 + R3 R5 R R 3 5 18 Circuits w/ Indep. Current Sources ❑ Grouping the node equations together, we obtain 1 1 1 1 1 − v 2 v 1 + + − v3 = iA R2 R3 R1 R2 R3 − v1 1 1 1 1 1 − v 3 + v 2 + + =0 R2 R R R R 2 4 5 5 − v1 ❑ 1 1 1 1 = − iB . − v2 + v 3 + R3 R5 R R 3 5 The equations can also be written in matrix form as 1 1 1 + + R R R3 2 1 1 − R2 1 − R3 Nodal and Loop Analysis Techniques 1 R2 1 1 1 + + R2 R4 R5 1 − R5 − 1 R3 1 − R5 1 1 + R3 R5 − v 1 i A v = 0 . 2 v 3 − iB 19 Circuits w/ Indep. Current Sources ❑ Symmetrical G matrix – The node equations for networks containing only resistors and independent current sources can always be written in symmetrical form. 1 1 + R R 2 1 − 1 R2 1 R2 1 1 + R21 R3 − 1 1 1 + + R R R3 2 1 1 − R2 1 − R3 Nodal and Loop Analysis Techniques 1 R2 1 1 1 + + R2 R4 R5 1 − R5 − 1 R3 1 − R5 1 1 + R3 R5 − 20 Circuits w/ Indep. Current Sources ❑ Symmetrical G matrix – We can take advantage of this fact and learn to write the equations by inspection. 1 1 + R R 2 1 − 1 R2 1 R2 v 1 i A = 1 1 v 2 − iB + R12 R3 − – 1st equation: The coefficient of v1 is the sum of all the conductances connected to node 1, the coefficient of v2 is the negative of the conductance connected between node 1 and node 2, and the right hand side is the sum of the currents entering node 1 through current sources. – 2nd equation: The coefficient of v2 is the sum of all the conductances connected to node 2, the coefficient of v1 is the negative of the conductance connected between node 2 and node 1, and the right-hand side is the sum of the currents entering node 2 through current sources. Nodal and Loop Analysis Techniques 21 Circuits w/ Indep. Current Sources ❑ Symmetrical G matrix 1st equation: The coefficient of v1 is the sum of all the conductances connected to node 1, the coefficient of v2 is the negative of the conductance connected between node 1 and node 2, the coefficient of v3 is the negative of the conductance connected between node 1 and node 3, and the right hand side is the sum of the currents entering node 1 through current sources. 1 1 1 + + R R R3 2 1 1 − R2 1 − R3 1 R2 1 1 1 + + R2 R4 R5 1 − R5 − Nodal and Loop Analysis Techniques 1 R3 v i 1 A 1 − v2 = 0 R5 1 1 v 3 − iB + R3 R5 − The other two equations are obtained in a similar manner. 22 Circuits w/ Indep. Current Sources ❑ For the node j with node voltage vj , – Coefficient of vj: sum of all the conductances connected to node j – Coefficients of the other node voltages (e.g., vj-1 , vj+1): negative of the sum of the conductances connected directly between these nodes and node j – Right-hand side of the equation: sum of the currents entering the node via current sources ❑ The left-hand side of the equation represents the sum of the currents leaving node j and the right-hand side represents the currents entering node j. Nodal and Loop Analysis Techniques 23 Example ❑ Determine the node voltages. 4 mA 2 k 4 0 − 2 1 1 1 − v 2 (0) − v 3 = − i A v 1 + R1 R2 R1 2 k 4 k 4 k 2 mA 1 k 0 − 2 v 1 − 16 2 − 1 v 2 = 8 −1 7 v 3 0 Nodal and Loop Analysis Techniques 1 1 1 − v 3 = i A − iB − v 1(0) + v 2 + R R R 4 3 4 − v 1( 1 1 1 1 1 = 0 + v 3 ) − v 2 + + R1 R R R R 4 1 4 5 1 1 + 2k 2k 0 − 1 2k 0 1 1 + 4k 4k 1 − 4k 1 v − 4m 2k 1 1 − v 2 = 2m 4k 1 1 1 v 3 0 + + 2k 4k 1k − 24 Example ❑ Let’s use Gaussian elimination to solve the equations. 4 0 − 2 0 − 2 v 1 − 16 2 − 1 v 2 = 8 −1 7 v 3 0 0 − 2 v 1 − 16 4 0 2 − 1 v 2 = 8 0 − 2 12 v 3 − 16 4 0 0 0 − 2 v 1 − 16 22 0 v 2 = 80 0 11 v 3 − 8 4 0 0 22 0 0 22 0 0 0 v 1 − 96 0 v 2 = 80 11 v 3 − 8 Nodal and Loop Analysis Techniques 0 − 2 v 1 − 16 2 − 1 v 2 = 8 0 11 v 3 − 8 1 0 0 v 1 − 4.3636 0 1 0 v = 3.6364 2 0 0 1 v 3 − 0.7273 25 Example ❑ Let’s use matrix analysis to solve the equations. We can write GV = I, where in this case 4 G= 0 − 2 0 − 2 v 1 − 16 2 − 1 , V = v 2 , and I = 8 . v 3 −1 7 0 The solution to the matrix equation is V = G−1I and therefore, v 1 4 v = 0 2 v 3 − 2 Nodal and Loop Analysis Techniques 0 − 2 2 − 1 −1 7 −1 − 16 8 0 26 Example ❑ Let’s use matrix analysis to solve the equations. To calculate the inverse of G, we use the following formula. 𝑎 A−1 = 𝑑 𝑔 𝑏 𝑒 ℎ 𝑐 𝑓 𝑖 −1 𝑒𝑖 − 𝑓ℎ 1 𝑐ℎ − 𝑏𝑖 = |A| 𝑏𝑓 − 𝑐𝑒 𝑓𝑔 − 𝑑𝑖 𝑎𝑖 − 𝑐𝑔 𝑐𝑑 − 𝑎𝑓 𝑑ℎ − 𝑒𝑔 𝑏𝑔 − 𝑎ℎ 𝑎𝑒 − 𝑏𝑑 | A | = a(ei − fh) + b(fg − di ) + c (dh − eg ) Therefore, | G | = 4 (2)(7) − (−1)(−1) + 0 (−1)(−2) − (0)(7) − 2 (0)(−1) − (2)(−2) = 44 v 1 13 2 v = 1 2 24 2 44 v 3 4 4 Nodal and Loop Analysis Techniques 4 − 16 − 192 − 4.3636 1 4 8 = 160 = 3.6363 44 8 0 − 32 − 0.7273 27 Example ❑ Let’s use MATLAB to solve the equations. 4 0 − 2 0 − 2 v 1 − 16 2 − 1 v 2 = 8 −1 7 v 3 0 >> G = [ 4 G = 4 0 -2 0 >> I = [ -16; I = -16 8 0 -2; 0 2 -1 8; 0 2 -1; -2 -1 7 ] -2 -1 7 0 ] >> V = inv(G)*I V = -4.3636 3.6364 -0.7273 Nodal and Loop Analysis Techniques 28 Summary ❑ Nodal Analysis – Value of knowing all the node voltages in a network – Circuits containing only independent current sources – Techniques to write node-voltage equations – Symmetry of node-voltage equations – Various examples – Techniques to solve node-voltage equations Nodal and Loop Analysis Techniques 29 Recommended Problems ❑ Problems 3.1.1, 3.1.2, 3.1.4, 3.1.6, 3.1.9 (in textbook) 30