Electric Circuits and Networks 10-09-2018 Lecture 16: Nodal/, Source Transformation Lecturer: Dr. Vinita Vasudevan Scribe: Shashank Shekhar

Nodal Analysis                    G  V  = Is       |{z} |{z} Unknown independent sources/ Variables initial conditions Mesh Analysis                    Z  I = Vs       |{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:

1 " # " #    1 V1 Is + Vs/2 2  = 1 1 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 −1   Iin (s) + Vc (0−) s V1           =    −    iL (0−)  1 1/s + 1 V2  −  s      0  Iin (s) Vc (0−)         =   +   +        i (0 )  0   0  − L −  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 source equal to zero ⇔ short the terminal Setting the 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 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