Electric Circuits and Networks 10-09-2018 Lecture 16: Nodal/Mesh Analysis, Source Transformation Lecturer: Dr. Vinita Vasudevan Scribe: Shashank Shekhar Nodal Analysis 2 3 2 3 2 3 6 7 6 7 6 7 6 7 6 7 6 7 6 G 7 6V 7 = 6Is7 46 57 46 57 46 57 |{z} |{z} Unknown independent sources= Variables initial conditions Mesh Analysis 2 3 2 3 2 3 6 7 6 7 6 7 6 7 6 7 6 7 6 Z 7 6I7 = 6Vs7 46 57 46 57 46 57 |{z} |{z} Unknown independent sources= Variables initial conditions Example 1: Write the nodal equations for following circuit super node 4i1 V1 V2 2 Ω V3 − + + Iin(s) 2 Ω 2 Ω Vs − ref As V3 = Vs so effectively we have only two unknown V1;V2. By applying KCL we will get following matrix equation: 21 3" # " # 6 7 6 17 V1 Is + Vs=2 62 7 = 41 15 V2 0 " # " # I V =2 = s + s 0 0 " # " # " # V1 1 Is 1 Vs=2 = G− +G− V2 0 0 | {z } | {z } S1 S2 Observation: Note that S can be obtained by setting V = 0,Circuit for that will be as 1 s follows: 1 V1 V2 2 Ω V3 − + 4i11 Iin (s) 2 Ω 2 Ω S2 can be obtained by setting Is = 0,Circuit for that will be as follows: V1 V2 2 Ω V3 − + 4i12 + 2 Ω 2 Ω Vs − and the total solution is sum of the two solutions. This is expected to be happen as the system is linear w.r.t. each source, so Superposition should be applicable. Example 2: Write the nodal equations for following circuit V1 1 Ω V1 iL (0−) Iin (s) 1 Ω 1F Vc (0−) 1H s 2 3 2 + 2 −1 32 3 6Iin (s) + Vc (0−)7 6s 76V17 6 7 6 76 7 6 7 6 76 7 = 6 7 46 − 5746 57 6 iL (0−) 7 1 1=s + 1 V2 4 − 5 s 2 3 2 3 2 0 3 Iin (s) Vc (0−) 6 7 6 7 6 7 6 7 = 6 7 + 6 7 + 6 7 6 7 6 7 6 i (0 )7 4 0 5 4 0 5 46− L − 57 s zero state Using superposition zero-input can apply one initial condition at a time 2 Example 3: Solve the following circuit using Superposition 3 Ω + 6 A 8 Ω − 2 V 3 Ω + 6 A 8 Ω 3 Ω 8 Ω − 2 V 3 2 i = 6 · i = 1 11 2 11 i = i1 + i2 Note: Superposition can be appiled only for independent sources and sources appearing because of initial conditions Setting the voltage source equal to zero , short the terminal Setting the current source equal to zero , opencircuit the terminal Source Transformation R s − V = Vs IRs + V + s Vs − V Voc - Vs equivalent Rs Isc + − Vin Vs Vs I = V Rs R Rs - Hence we can replace voltage source with resistance in series by a current source with resis- tance in parallel. 3 Example 4: Replace current source with resistance in parallel by voltage source with resistance in series. R2 R3 + − Vx + Is R1 Vs − Since the following two circuits are equivalent Rs + + () + Is R1 IsR1 Vs − V - - So the given circuit can be replaced by following: R1 R2 R3 − + − V IsR1 Vs x I = R1 + R2 + R3 + + IsR1 − I Vs − Vx = R3I Note : Source transformation can be used for independent, dependent sources and also for initial conditions, as the I-V relationship is the same in all conditions Exercise 1 : Apply the series of source transformations to get circuit with single mesh 5 Ω 3 Ω 4 Ω + − Vx 5Vx + 3 V − 7 Ω 9 Ω 9 A 4 Exercise 2 : Replace the dashed part of given circuit by single voltage source in series with a inpedence 20 Ω 14 Ω − 12 V (s) + 8s Z (s) in s 5.