* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download I 2 - s3.amazonaws.com
Survey
Document related concepts
Three-phase electric power wikipedia , lookup
Switched-mode power supply wikipedia , lookup
Surge protector wikipedia , lookup
Buck converter wikipedia , lookup
Resistive opto-isolator wikipedia , lookup
Voltage optimisation wikipedia , lookup
Rectiverter wikipedia , lookup
Current source wikipedia , lookup
Mathematics of radio engineering wikipedia , lookup
Signal-flow graph wikipedia , lookup
Opto-isolator wikipedia , lookup
Stray voltage wikipedia , lookup
Topology (electrical circuits) wikipedia , lookup
Ground loop (electricity) wikipedia , lookup
Alternating current wikipedia , lookup
Mains electricity wikipedia , lookup
Transcript
ECE 3144 Lecture 14 Dr. Rose Q. Hu Electrical and Computer Engineering Department Mississippi State University 1 Loop analysis • In loop analysis, the unknown parameters are currents and KVL is employed to determine the unknowns. • If there are N independent loops, N independent equations are needed to describe the network. 2 Nodal analysis vs. loop analysis We would like to consider nodal analysis and loop analysis as mirror scenarios: • In nodal analysis, the unknown parameters are nodal voltages and KCL is used to solve the problem. • If there are N nodes in the network, N-1 equations are required to solve the problem. • KCL is applied to each nonreference node. • In loop analysis, the unknown parameters are current and KVL is employed to determine the unknowns. • If there are N independent loops, N independent equations are needed to describe the network. • KVL is applied to each independent loop. 3 Loop analysis case 1: containing independent voltage source only vS1 a 2 kW b vS2 + d + 12 V 2 kW v1 c 2 kW v2 - I1 4V I2 e •Let us first identify how many independent loops here: + Loop 1 and loop 2 v3 •The unknown variables are loop currents: I1 and I2 - •Now we determine the physical currents through each branch. f Branch d->a: I1 Branch a->b: I1 Branch b->c: I2 Branch c->f: I2 Branch f->e: I2 Branch b->e: I1-I2 Applying KVL to loop 1: v1 – vS1 + v2 = 0 => 2000I1 + 2000(I1-I2) -12 = 0 => 4000I1 - 2000I2 = 12 => -2000 (I1-I2) + 4 + 2000I2 = 0 (1) Applying KVL to loop 2: -v2 + vS2 + v3= 0 => -2000I1 + 4000I2 = -4 (2) There are two equations and two unknowns I1 and I2. You can solve them. I1 =10/3 mA and I2 =2/3 mA 4 Loop analysis case 1: cont’d The two equations are put into the matrix format: 4000 2000 2000 I1 12 4000 I 2 4 Again, we notice that the coefficient matrix for this type of circuits is symmetrical. We will explain later why this coefficient matrix is symmetrical and how to write the mesh equations by inspection. An important definition: what is mesh: A mesh is a special kind of loop that does not contain any loops within it. So when you traverse the path of a mesh, you do not encircle any circuit elements. The majority of our loop analysis will involve writing KVL equations for meshes, thus loop analysis usually is called mesh analysis. The following example derives the general interpretation on the symmetry of the coefficient matrix. 5 Loop analysis case 1: independent voltage source only There are two independent loops (meshes) in this circuit. Applying KVL to loop 1 v1 + v3 + v2 – vS1 = 0 => i1R1 + (i1-i2)R3 + i1R2 - vS1 = 0 i1(R1 + R2 + R3) - i2R3 =vS1 => (1) Applying KVL to loop 2: v4 + v5 –v3 + vS2 = 0 => i2R4 + i2R5 – (i1-i2)R3 + vS2 = 0 -i1R3 +i2(R3 + R4 + R5) =-vS2 In the matrix format => (2) R3 R1 R2 R3 i1 vs1 i 2 vs2 R 3 R 3 R 4 R 5 It is again the Ohm’s law in matrix format: RI = V. 6 Remember in nodal analysis case 1 we have Ohms' law in matrix format GV = I Loop analysis case 1: independent voltage source only • • • • the R matrix is symmetrical. In the first equation, the coefficient of i1 is the sum of all resistors through which mesh current i1 flows; the coefficient of i2 is the negative of the sum of the resistances common to mesh current 1 and mesh current 2. The right-hand side of the equation is the algebraic sum of the voltages sources in mesh 1. The sign of the voltage source is positive if it aids the assumed direction of current flow 1 and negative if it apposes the assumed flow direction. In the second equation, the coefficient of i2 is the sum of all resistors through which mesh current i2 flows; the coefficient of i1 is the negative of the sum of the resistances common to mesh current 1 and mesh current 2. The right-hand side of the equation is the algebraic sum of the voltages sources in mesh 2. The sign of the voltage source is positive if it aids the assumed direction of current flow 2 and negative if it apposes the assumed flow direction. In general, we assume all of the mesh currents to be in the same direction (clockwise or counterclockwise). If KVL is applied to mesh j with mesh current ij, the coefficient of ij is the sum of all resistors in mesh j; the coefficients for other mesh currents ik (kj) are the negative sum of the resistors common to mesh k and mesh j.The right-hand side of jth equation is equal to the algebraic sum of the voltage sources in mesh j. These voltage sources have a positive sign if they aid the current flow ij and a negative sum if they appose it. 7 Loop analysis case 2: with dependent voltage source We deal with circuits with dependent voltage source the same way as the independent voltage source except that we need to write the controlling equations for the dependent voltage source. However the symmetry of the R matrix may no longer exist. In loop 1, apply KVL: 2 kW 2000 Ix I1 2000 I x 2 kW I1 4 kW I1 I 2 0 + 6V 4 kW I2 2 kW Ix Vo (1) In loop 2, apply KVL 6 V 2 kW I2 4 kW I 2 I1 0 (2) Write the controlling equation for the dependent voltage source I x I2 (3) There are three equations for three unknowns I1, I2, Ix. I1 =3 mA I2 = 3 mA Ix = 3 mA 8 Homework for lecture 14 • Problem 3.43, 3.45, 3.48, 3.51 • Due Feb 18 9