Nodal & Loop Analysis, Schemes and Mind Maps of Law

Download Nodal & Loop Analysis and more Schemes and Mind Maps Law in PDF only on Docsity! ELT221 1 Nodal & Loop Analysis - Nodal Analysis - Loop Analysis - Mesh Analysis Enzo Paterno EP Revision 12/9/14 ELT221 2 NODAL ANALYSIS Enzo Paterno 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 (i.e. ground) and assumed to be positive voltages. For example the circuit below has four node voltages, (VS , Va , Vb , Vc ) with respect to the ground node. Once the node voltages in a circuit are defined , we can calculate the current through any resistive element using Ohm’s law: R VV I NN −+ − = VS Va Vb Vc + - I R +  ELT221 5 NODAL ANALYSIS Enzo Paterno 3 21 101 6 1 4 1 −= − + xV k V k 3 21 104 3 1 6 1 −= − + xV k V k We have a 2x2 linear system. Solve for V1 and V2. How do we solve an n x n linear system? TI-83 Plus  Method of substitution  Method of elimination  Gaussian elimination  Cramer’s rule  Matrix SW calculator  MATLAB software tool  JAVA Applet ELT221 Enzo Paterno 6 GAUSSIAN ELIMINATION Two simultaneous linear equations form a 2x2 linear system: This 2x2 linear system can be represented in matrix form: [ ][ ] [ ]KxM K K DC BA x x =→      =            2 1 2 1 Using matrix theory, the solution to this linear system can be found by: [M]-1 is the inverse of matrix [M] M is a square matrix, x and K are column matrices (vectors) A x1 + B x2 = K1 C x1 + D x2 = K2 [ ] [ ] [ ]KMx 1− = ELT221 Enzo Paterno 7 Below is a general system of m equations with n unknowns: This system can be expressed as a matrix equation: Ax = b The system solution is given by: x = A-1 b GAUSSIAN ELIMINATION ELT221 10 NODAL ANALYSIS – CRAMER’S RULE Enzo Paterno V kk kk k x k x V 6 3 1 6 1 6 1 4 1 3 1104 6 1101 3 3 1 −= − − − − = − − 3 21 101 6 1 4 1 −= − + xV k V k 3 21 104 3 1 6 1 −= − + xV k V k V kk kk x k x k V 15 3 1 6 1 6 1 4 1 104 6 1 101 4 1 3 3 2 −= − − = − − ELT221 11 NODAL ANALYSIS Enzo Paterno SV aV bV cV 0 :@ 321 321 =++− += III IIIVa 0 :@ 543 543 =++− += III IIIVb k V kkk V k V k VV k V k VV c ba cbbab 99 1 4 1 3 1 3 1 0 943 =    +++− = − ++ − k V k V kkk V k VV k V k VV sb a baasa 933 1 6 1 9 1 0 369 =−    ++ = − ++ − Vs is a constant Vs = 12 v 0 3 1 9 1 9 1 39 :@ =    +− = − kk V k V k V k VVV cb ccb C ELT221 12 NODAL ANALYSIS Enzo Paterno VV VV VVVVV k VV k V k VV 5.1 6)6(124 02 0 12612 2 2 32212 32212 = =−+= =−++− = − ++ − VV VV 6 12 3 1 −= = V1 =12 V V3 = -6 V Only one KCL needed ELT221 15 NODAL ANALYSIS Enzo Paterno V2 I3 I4 I5 V1 V3 06134:2 46644 846 :@ 321 23221 23221 5432 =+− +−=− + − = − += VVVEq VVVVV k V k VV k VV IIIV ELT221 16 NODAL ANALYSIS Enzo Paterno V1 V2 V3 I2 I4 I6 042:3 22 884 :@ 321 33132 33132 6243 =−+ =−+− = − + − =+ VVVEq VVVVV k V k VV k VV IIIV ELT221 NODAL ANALYSIS 17 Enzo Paterno           =                     − − −− 0 0 72 421 6134 3413 3 2 1 V V V                     − − −− =           − 0 0 72 421 6134 3413 1 3 2 1 V V V 723413:1 321 =−− VVVEq 06134:2 321 =+− VVVEq 042:3 321 =−+ VVVEq 32 VVVo −= ELT221 20 Enzo Paterno 2 2 1 21 2 21 1 1 211 111 :@ v R v RR I R vv R vI IIIV A A A −      += − += += 2 32 1 2 2 32 21 32322 111 1 :@ v RR v R I v RR vvI IIIIIIV A B BB       +−−= − − = −=⇒+= 3 21 101 6 1 6 1 12 1 −=−      + xv k v kk 3 21 104 6 1 6 1 6 1 −−=      +−− xv kk v k R1 = 12 kΩ, R2 = 6 kΩ, R3 = 6 kΩ NODAL ANALYSIS – Circuits with independent current sources ELT221 21 Enzo Paterno       −      − − =      − −− −− −− 3 31 33 33 2 1 104 101 1033.1016. 1016.1025. x x xx xx V V NODAL ANALYSIS – Circuits with independent current sources ELT221 22 Enzo Paterno Inverse -6 -15 NODAL ANALYSIS – Circuits with independent current sources ELT221 25 MESH LOOP ANALYSIS Enzo Paterno In a loop analysis, the variables in the circuit are selected to be the branch currents and KVL equations are used so that the variables are the unknown currents. The number of LI KVL equations necessary to determine all the currents in a network with B branches, N nodes is: B – N + 1. For example the circuit below requires 7 – 6 + 1 = 2 LI KVL. We identify two independent mesh loops ABEF and BCDE. A mesh is a loop that does not contain any other loop within it. ELT221 26 Enzo Paterno ( ) 1323211 132312111 213231111 0 :@ S S S vRiRRRi vRiRiRiRi RiRiRiRiv ABEF −=+++− −=+−−− =−+−− MESH LOOP ANALYSIS ( ) 2543231 232524231 313252422 0 :@ S S S vRRRiRi vRiRiRiRi RiRiRiRiv BCDE =++−+ =−−−+ =+−−−− - + ( ) ( ) ( ) ( )      − =            ++− ++− =++−+ −=+++− 2 1 2 1 5433 3321 2543231 1323211 S S S S v v i i RRRR RRRR vRRRiRi vRiRRRi + - ELT221 27 Enzo Paterno MESH LOOP ANALYSIS 12612 066612 21 211 =− =+−− ikik ikikik + + 396 03663 21 212 =− =−+−− ikik ikikik 396 12612 21 21 =− =− ikik ikik 61812 12612 21 21 −=+− =− ikik ikik 612 2 =ik mA k i 5.0 12 6 2 == mA k i 25.1 12 15 1 == Use Mesh analysis to solve for I1 and I2. Ss MESH LOOP ANALYSIS ELT221 > PSPICE Simulation R1 _R2 * 6k Enzo Paterno 30 ELT221 31 Enzo Paterno MESH LOOP ANALYSIS i1 i2 Use Mesh analysis to solve for I1 and I2. + + + + + Mesh1 Mesh2 V + - 10055.12 0555.7100 21 211 =−+ =+−− ikik ikikik 0105 05523 21 1222 =−+ =+−−− ikik ikikikik Mesh1: Mesh2: ELT221 32 Enzo Paterno MESH LOOP ANALYSIS 0105 10055.12 21 21 =−+ =−+ ikik ikik mA kk kk k k i 10 105 55.12 100 5100 1 = − − − − = We use Cramer’s rule to find the currents i1 and i2: Remark: This step could have been solved using Elimination by addition or Elimination by substitution techniques bcad dc ba −= 2x2 determinant mA kk kk k k i 5 105 55.12 05 1005.12 2 = − − = ELT221 35 Enzo Paterno MESH LOOP ANALYSIS 21 2 1 2 5 2 5 1 31 12 1 2 VV V V i += − − − = VOUT = 2 i2. We use Cramer’s rule to find the current i2: 21232 5 4 5 22 VViRiVOUT +===∴ 221 121 3 2 Vii Vii =+− =−+ If V1 = V2 = 10 v then VOUT = 4 + 8 = 12 V ELT221 36 Enzo Paterno MESH LOOP ANALYSIS Use Mesh analysis to solve for I1, I2, and I3. 6610 06646 31 311 =+− =+−−− ikik ikikik + + + + 6312 09336 32 232 =−+ =−+−+ ikik ikikik 02136 0331266 321 23313 =−++ =+−−+− ikikik ikikikikik           =                     − − − 0 6 6 2136 3120 6010 3 2 1 i i i kkk kk kk                       − − − =           − 0 6 6 2136 3120 6010 1 3 2 1 k k i i i 1: 2: 3: ELT221 37 Enzo Paterno MESH LOOP ANALYSIS                       − − − =           − 0 6 6 2136 3120 6010 1 3 2 1 k k i i i ELT221 40 Enzo Paterno MESH LOOP ANALYSIS There are three mesh currents. I1 and I2 are known. Only one mesh equation is needed for I3. I1 = 4 mA I2 = -2 mA mA k i ik ikikik ikikikikik 25.0 12 3 31288 31242 0622443 3 3 321 31323 == =++− =+−− =−+−+−+ + + + + Use Mesh analysis to solve Vo. Vo = - [0.25 mA (6kΩ) ] + 3 = 1.5 v ELT221 41 Enzo Paterno MESH LOOP ANALYSIS Use Mesh analysis to solve for I1, I2, and I3. 12 122 13 133 24 0222 126 01116 ikikV ikikikV ikikV ikikVik x x x x −= =+−− +−+= =+−−−+ + + k ikiki ikikikik ikikVikik x 2 634 61242 24126 12 3 1123 1213 − −− = −−−=− −==+−+ + + @3: @2: I2 = (4 - 2/3) = 3.33 mA One could have used SuperMesh (take out the 4 mA ) Eq1: Eq2: Eq1 Eq2 I1 = 2 mA We see that 4mA = I2 – I3 mAi k iki k imAki 3 2 2 44 2 66)4(4 3 3 3 3 3 − =⇒ − + = − −−+ =