EE-229

Departmental Lab 1

Circuits Lab

Performance Report

Submitted To: Submitted By:

Professor S. K. Suman Viresh Verma

SID: 12104085

B.E. (4th Semester)

Electrical Engineering

Department

CONTENTS

Network Analysis using MATLAB

To verify Thevenins Theorem using MATLAB

To verify Nortons Theorem using MATLAB

To verify Superposition Theorem using MATLAB

To verify Millmans Theorem using MATLAB

To verify Maximum Power Transfer Theorem

using MATLAB

To verify the Voltage and Current

relationships in Three Phase Circuits using

MATLAB

Case 1: Wye Connected Supply, Wye

Connected Load, 4 Wire Connections

Case 2: Wye Connected Supply, Wye

Connected Load, 3 Wire Connections

Case 3: Delta Connected Supply,

Delta Connected Load

Case 4: Wye Connected Supply, Delta

Connected Load

Case 5: Delta Connected Supply, Wye

Connected Load

To find out the Z, Y, G, H and ABCD

Parameters for a Two-Port Network(TPN) using

MATLAB

Case 1: Z Parameters

Case 2: Y Parameters

Case 3: G Parameters

Case 4: H Parameters

Case 5: ABCD Parameters

Here

V (Thevenin) = V (O.C.)=32.39/_24.89V

I (S.C.)=2.712/_0.6888A

Z (Thevenin) = V (O.C.)/I (S.C.)=11.9432/_24.2012=10.8935+4.896j

Here

I (Norton) = I (S.C.)=2.712/_0.6888A

Z (Norton) = V(O.C.)/I(S.C.)=11.9432/_24.2012=10.8935+4.896j

I(Load)=5.46/_30.34

I (1) =5.948/_31.05

I (2) =1.497/_178

I(1)+I(2)=5.45/_30.40 ~ I(Load)

V1 =100/_0V V2=90/_45V V3=80/_30V

Z1= 5+8j Z2=10-6j Z3=20+12j-4j

V (Millman) = (V1/Z1)+(V2/Z2)+(V3/Z3)=60.93/_15.057V

Z (Millman)-1= Z1-1+Z2-1+Z3-1

Here

V (Thevenin) =V (O.C.)=32.39/_24.89 V

I(S.C.)=2.712/_0.6888 A

Z (Thevenin) =V(O.C.)/I(S.C.)=11.

The power transferred is maximum when the load impedance is the same as the Thevenin

impedance

CASE 1: WYE CONNECTED SUPPLY, WYE CONNECTED LOAD, 4 WIRE CONNECTIONS,

BALANCED AND UNBALANCED LOAD

Here

|I (OA)|=|I (OB)|=|I (OC)|

I (OA), I (OB) and I (OC) are all separated by an angle of 120 degrees.

|V (OA)|=|V (OB)|=|V (OC)|

V (OA), V (OB) and V (OC) are all separated by an angle of 120 degrees.

Magnitude of Neutral Current, I (NO), is of the order of 10-15, which shows the current flowing

through the neutral is almost zero. This must be true because this is a Balanced Load.

Here

|I (NA)| |I (NB)| |I (NC)|

|V (NA)|=|V (NB)|=|V (NC)|

V (NA), V (NB) and V (NC) are all separated by an angle of 120 degrees.

Magnitude of Neutral Current, I (NN)is finite which shows there is current flowing through the

neutral. This must be true because this is an Unbalanced Load.

CASE 2: WYE CONNECTED SUPPLY, WYE CONNECTED LOAD, 3 WIRE

CONNECTIONS, BALANCED AND UNBALANCED LOAD

Here

|I (OA)|=|I (OB)|=|I (OC)|

I (OA), I (OB) and I (OC) are all separated by an angle of 120 degrees.

|V (OA)|=|V (OB)|=|V (OC)|

V (OA), V (OB) and V (OC) are all separated by an angle of 120 degrees.

Here

|I (OA)| |I (OB)| |I (OC)|

I (OA), I (OB) and I (OC) are not separated by an angle of 120 degrees.

|V (OA)| |V (OB)| |V (OC)|

V (OA), V (OB) and V (OC) are not separated by an angle of 120 degrees.

Magnitude of Neutral Current, I (NO), is of the order of 10-15, which shows the current flowing

through the neutral is almost zero. This must be true because this is a Balanced Load.

CASE 3: DELTA CONNECTED SUPPLY, DELTA CONNECTED LOAD, BALANCED AND

UNBALANCED LOAD

Here

|I (AB)|=|I (BC)|=|I (CA)|

|V (OA)|=|V (OB)|=|V (OC)|

|I (AB)| |I (BC)| |I (CA)|

|V (OA)| |V (OB)| |V (OC)|

The behavior is on expected lines as in the first case; the load was Balanced while it was Unbalanced

in the second case.

CASE 4: WYE CONNECTED SUPPLY, DELTA CONNECTED LOAD, BALANCED AND

UNBALANCED LOAD

Here

|I (AA)|=|I (BB)|=|I (CC)|

I (AA), I (BB) and I (CC) are all separated by an angle of 120 degrees.

|V (AB)|=|V (BC)|=|V (CA)|

V (AB), V (BC) and V (CA) are all separated by an angle of 120 degrees.

This is because the Load is Balanced.

Here

|I (OA)| |I (OB)| |I (OC)|

I (OA), I (OB) and I (OC) are not separated by an angle of 120 degrees.

|V (OA)| |V (OB)| |V (OC)|

V (OA), V (OB) and V (OC) are not separated by an angle of 120 degrees.

CASE 5: DELTA CONNECTED SUPPLY, WYE CONNECTED LOAD, BALNCED AND

UNBALANCED LOAD

Here

|I (AA)|=|I (BB)|=|I (CC)|

I (AA), I (BB) and I (CC) are all separated by an angle of 120 degrees.

|V (OA)|=|V (OB)|=|V (OC)|

V (OA), V (OB) and V (OC) are all separated by an angle of 120 degrees.

This is because the Load is Balanced.

Here

|I (AA)| |I (BB)| |I (CC)|

I (AA), I (BB) and I (CC) are not separated by an angle of 120 degrees.

|V (OA)| |V (OB)| |V (OC)|

V (OA), V (OB) and V (OC) are not separated by an angle of 120 degrees.

This is because the Load is Unbalanced.

CASE 1: Z PARAMETERS

CASE 2: Y PARAMETERS

CASE 3: G PARAMETERS

CASE 4: H PARAMETERS

CASE 5: ABCD PARAMETERS