EE201 Circuit Theory Ch 3. Network Theorems (Part 3) 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 Loop Analysis ❑ A loop analysis uses KVL to determine a set of loop currents in the circuit. – Once these loop currents are known, Ohm’s law can be used to calculate any voltages in the network. – Via network topology it can be shown that, in general, there are exactly B − N + 1 linearly independent KVL equations for any network, where B is the number of branches in the circuit and N is the number of nodes. * Loop: a closed path that does not go twice over any node. 3 Loop Analysis ❑ As an example, consider the circuit shown below. – There are 8 branches and 5 nodes. – Thus, the number of independent KVL equations necessary to determine all currents in the network is B − N + 1 = 8 − 5 + 1 = 4. If we define 4 loop currents as shown in the figure, the branch currents are then determined as i1(t) = iA(t) i2(t) = iA(t) − iB(t) i3(t) = iB(t) i4(t) = iA(t) − iC(t) i5(t) = iB(t) − iD(t) i6(t) = −iC(t) i7(t) = iC(t) − iD(t) i8(t) = −iD(t). 4 Circuits w/ Indep. Voltage Sources ❑ Consider the network shown below. – This network has 7 branches and 6 nodes, and thus the number of linearly independent KVL equations necessary to determine all currents in the circuit is B − N + 1 = 7 − 6 + 1 = 2. – Thus, we identify 2 independent loops, A-B-E-F-A and B-C-D-E-B. – We now define loop currents, which can be used to find the physical currents in the circuit. – Let us assume that i1 flows in the 1st loop and i2 flows in the 2nd loop. – Then, the branch current flowing from B to E through R3 is i1 − i2. 5 Circuits w/ Indep. Voltage Sources ❑ Consider the network shown below. – Applying KVL to the 1st loop yields +v1 + v3 + v2 − vS1 = 0. – KVL applied to the 2nd loop yields +vS2 + v4 + v5 − v3 = 0. – Substituting v1 = i1R1, v2 = i1R2, v3 = (i1 − i2)R3, v4 = i2R4, and v5 = i2R5 into the two KVL equations produces i1(R1 + R2 + R3) − i2(R3) = vs1, and −i1(R3) + i2(R3 + R4 + R5) = −vs2. 6 Circuits w/ Indep. Voltage Sources ❑ Consider the network shown below. i1(R1 + R2 + R3) − i2(R3) = vs1, and −i1(R3) + i2(R3 + R4 + R5) = −vs2. – Rewriting the derived mesh equations in matrix form, we obtain R1 + R2 + R3 − R3 − R3 i1 v S 1 = . R3 + R4 + R5 i2 − v S 2 7 Mesh Current & Mesh Analysis ❑ Consider the network shown below. – A mesh is a special kind of loop that does not contain any loops within it. – Therefore, as we traverse the path of a mesh, we do not encircle any circuit elements. – For example, the circuit contains two meshes defined by the paths A-B-E-FA and B-C-D-E-B. The path A-B-C-D-E-F-A is a loop, but not a mesh. – A mesh current is a loop current associated with a mesh. – A mesh analysis deals with mesh currents as unknowns. 8 Example: Mesh Analysis ❑ Find the current Io. Results from two simultaneous equations: I1 = 54 mA, I2 = 12 mA Io = 34 mA Since we know the branch currents, – Mesh equations: −12 + 6kI1 + 6k(I1 − I2) = 0 6k(I2 − I1) + 3kI2 + 3 = 0 where Io = I1 − I2. we can use KVL around any closed path to check the results. 9 Example: Loop Analysis ❑ Find the current Io. Loop equations: −12 + 6k(I1 + I2) + 6kI1 = 0 −12 + 6k(I1 + I2) + 3kI2 + 3 = 0 Results from two simultaneous equations: – Since we want to calculate the current Io, we could use different loops for analysis. – Now the loop current I1 passes through the center leg of the network and I1 = Io. I1 = Io = 34 mA, I2 = 12 mA Current in the 12-V source = I1 + I2 = 54 mA These results agree with the mesh analysis. 10 Example: Nodal Analysis ❑ Find the current Io. Results from the node-voltage equation: Vo = 92 V Io = Vo6k = 34 mA In this case, we had to solve only one equation instead of two. – We could also use nodal analysis to find Io. – Node-voltage equation: Vo − 12 Vo Vo − 3 + + =0 6k 6k 3k 11 Symmetry of Mesh Equations R1 + R2 + R3 − R3 − R3 i1 v S 1 = R3 + R4 + R5 i2 − v S 2 The coefficient matrix for circuits containing resistors and independent voltage sources shows symmetry. ❑ We can write the mesh equations by inspection. ❑ – 1st equation: coefficient of i1 = (resistances in mesh 1); coefficient of i2 = − (resistances common to mesh 1 and mesh 2); right-hand side = (voltage sources in mesh 1) – 2nd equation: coefficient of i2 = (resistances in mesh 2); coefficient of i1 = − (resistances common to mesh 1 and mesh 2); right-hand side = (voltage sources in mesh 2) 12 Example ❑ Write the mesh equations by inspection. Matrix form of the equations: 0 − 6k I1 − 6 10k 0 12k − 3k I = 6 2 − 6k − 3k 21k I3 0 Dividing by 1000 yields – Mesh equations: (4k + 6k)I1 − (0)I2 − (6k)I3 = −6 −(0)I1 + (9k + 3k)I2 − (3k)I3 = 6 −(6k)I1 − (3k)I2 + (3k + 6k + 12k)I3 = 0 0 − 6 I1 − 0.006 10 0 12 − 3 I = 0.006 . 2 − 6 − 3 21 I3 0 13 Circuits w/ Indep. Current Sources As the presence of a voltage source simplified the nodal analysis, the presence of a current source simplifies a loop analysis. ❑ Let us consider the circuit shown below. ❑ – Although it appears that there are two unknown mesh currents, I1 goes directly through the current source and therefore is constrained to be 2 mA. Hence, only I2 is unknown. – Applying KVL to the rightmost mesh gives 2k(I2 − I1) − 2 + 6kI2 = 0. – Solving the mesh equation with knowing that I1 = 2 mA, I2 = 34 mA. 14 Example: Unshared Sources ❑ Find Vo. 1 mesh equation: 4k(I3 − I2) + 2k(I3 − I1) + 6kI3 − 3 = 0 By solving these three equations, we obtain I3 = 14 mA and Vo = 6kI3 − 3 = −32 V. – 3 independent equations are required. – 2 constraint equations: I1 = 4 mA I2 = −2 mA 15 Example: Shared Sources ❑ Find Io. The remaining loop current I3 must pass through the circuit elements not covered by the 2 previous equations, and cannot pass through the current sources. The path for this remaining loop current can be found by opencircuiting the current sources. – First, we select two loop currents I1 and I2 such that I1 and I2 pass directly through the 2-mA and 4mA sources, respectively. – Therefore, I1 = 2 mA and I2 = 4 mA. 16 Example: Shared Sources ❑ Find Io. – The KVL equation for this last loop is −6 + 1kI3 + 2k(I2 + I3) + 2k(I3 + I2 − I1) + 1k(I3 − I1) = 0. – Solving the equations yields I3 = −23 mA and therefore Io = I1 − I2 − I3 = −43 mA. 17 Example: Use of Supermesh ❑ Find Io. – 3 mesh currents are specified as shown below. – The unknown voltage across the 4-mA current source is assumed to be Vx. – Constraint equations: I1 = 2 mA and I2 − I3 = 4 mA. KVL equations for meshes 2 and 3: 2kI2 + 2k(I2 − I1) − Vx = 0 −6 + 1kI3 + Vx + 1k(I3 − I1) = 0 Adding these two equations yields −6 + 1kI3 + 2kI2 + 2k(I2 − I1) + 1k(I3 − I1) = 0 The unknown voltage Vx has been eliminated. 2 constraint equations together with this latter equation, yield the desired result. 18 Example: Use of Supermesh ❑ Find Io. – The supermesh approach can be used to avoid introducing the unknown voltage Vx. – The supermesh is created by mentally removing the 4-mA current source. Writing the KVL equation around the dotted path, which defines the supermesh, yields −6 + 1kI3 + 2kI2 + 2k(I2 − I1) + 1k(I3 − I1) = 0 This supermesh equation is the same as that obtained earlier by introducing the voltage Vx. 19 Circuits w/ Dependent Sources First, we treat the dependent source as if it were an independent source when writing the KVL equations. ❑ Then, we write the controlling equation for the dependent source. ❑ Let us consider the circuit shown below. ❑ Controlling equation: Vx = 4kI1 These equations can be combined to produce −2kI1 + 2kI2 = 0 – Loop-current equations: −2Vx + 2k(I1 + I2) + 4kI1 = 0 −2Vx + 2k(I1 + I2) − 3 + 6kI2 = 0 −6kI1 + 8kI2 = 3. In matrix form, the equations are − 2000 2000 I1 0 − 6000 8000 I = 3 . 2 20 Example ❑ Derive loop-current equations. Controlling equations: Vx = 2k(I3 − I1) Ix = I4 − I2 Combining these equations yields I1 = 4k I1 + I2 − I3 = 0 1kI2 + 3kI3 − 2kI4 = 8 – Constraint equations: I1 = 4k I2 = Vx2k – Mesh equations: −1kIx + 2k(I3 − I1) + 1k(I3 − I4) = 0 1k(I4 − I3) + 1k(I4 − I2) + 12 = 0 1kI2 + 1kI3 − 2kI4 = 12. In matrix form, the equations are 0 0 0 I1 0.004 1 1 1 −1 0 I 2 0 = . 0 1000 3000 − 2000 I3 8 0 1000 1000 − 2000 I 4 12 21 Example ❑ Find Vo in the network using nodal analysis. KCL equation at the supernode: − 2 V x V x − V3 3V x − V3 3V x − 4 + + + + =0 k 1k 1k 1k 1k KCL equation at the node labeled V3: − − 4V x + 2V3 = 2 V4 = 4 → V1 = 3Vx Simplified equations: 8V x − 2V3 = 6 Constraint equations: V1 − V2 = 2 Vx where Vx = V2 2 V3 − 3V x V3 − V x + + =0 k 1k 1k Solving these equations, we obtain Vx = 2 V, V3 = 5 V, and Vo = 3Vx − V3 = 1 V. 22 Example ❑ Find Vo in the network using loop analysis. KVL equations for meshes 2 and 4: −2Vx + 1kI2 + 1k(I2 − I3) = 0 1k(I4 + I3 − I1) − 2Vx + 1kI4 + 4 = 0 Controlling equation: Vx = 1k(I1 − I3 − I4) Substituting the equations for I1 and I3 into the 2 KVL equations yields The network has 4 loops, and thus 4 linearly independent equations are required. Constraint equations: I1 = 2k 2kI2 + 2kI4 = 6 4kI4 = 8. Solving these equations for I2 and I4, we obtain I4 = 2 mA and I2 = 1 mA, leading to Vo = 1 V. I3 = −2k 23 Nodal Analysis vs. Loop Analysis ❑ It is important to examine the circuit carefully before selecting an analysis approach. – One method could be much simpler than another, and a little time invested up front may save a lot of time in the long run. ❑ One consideration in the selection of a method should be the number of linearly independent equations that must be formulated. – For an N-node circuit, N − 1 linearly independent equations must be formulated to solve for N − 1 node voltages. – An N-loop circuit requires the formulation of N linearly independent equations. 24 Nodal Analysis vs. Loop Analysis ❑ In the previous example, the same circuit was solved using both node analysis and loop analysis. – The circuit has four unknown node voltages, which requires four linearly independent equations. Because there are two voltage sources, two constraint equations are needed. – This same circuit has four loops, which requires four linearly independent equations.The two current sources produce two constraint equations. The effort required to solve this circuit using either nodal or loop analysis is similar. However, this is not true for many circuits. 25 Nodal Analysis vs. Loop Analysis ❑ Consider the circuit shown below. – This circuit has 8 loops. – Selection of the loop currents such that only 1 loop current flows through the independent current source leaves us with 7 unknown loop currents. – This circuit has 7 nodes. – By selecting the bottom node as the reference node, 4 of the node voltages are known, leaving just 2 unknown node voltages−the node voltage across the current source and the node voltage across the 3- and 6- resistors. – Applying KCL at these 2 nodes yields 2 equations that can be solved for 2 unknown node voltages. 26 Loop Analysis: Strategy ❑ Determine the number of independent loops in the circuit. ❑ Assign a loop current to each independent loop. – For an N-loop circuit, there are N loop currents, requiring N linearly independent equations to solve for the loop currents. ❑ If any current sources in the circuit, either of two techniques can be employed. – In the first case, one loop current is selected to pass through one of the current sources. The remaining loop currents are determined in such a way that they don’t flow through the current source. – In the second case, a current is assigned to each mesh in the circuit and use super mesh approach. 27 Loop Analysis: Strategy ❑ Write a constraint equation for each current source−independent or dependent−in terms of the assigned loop current using KCL. – NI current sources yield NI linearly independent equations. – For each dependent current source, express the controlling variable for that source in terms of the loop currents. ❑ Use KVL to formulate the remaining N − NI linearly independent equations. ❑ Treat dependent voltage sources like independent voltage sources when formulating the KVL equations. – For each dependent voltage source, express the controlling variable in terms of the loop currents. 28 Summary ❑ Loop Analysis – Basic concepts – Planar circuits – Circuits containing only independent voltage sources – Mesh currents and mesh analysis – Symmetry of mesh equations – Circuits containing independent current sources – Supermesh approach – Circuits containing dependent sources – Nodal analysis vs. loop analysis – Strategy of loop analysis 29 Recommended Problems ❑ Problems 3.2.1, 3.2.5, 3.2.12, 3.2.14, 3.2.22, 3.2.26, 3.2.30, 3.2.35, 3.2.38 Nodal and Loop Analysis Techniques 30