EXPERIMENT NUMBER:01 Date of Performance : Date of ...dmce.ac.in/newdmcewebsite/others/resource/Humanities/BEE MANUAL18-19.pdf · PROCEDURE: Make the connections as shown in the diagram

EXPERIMENT NUMBER:01

Date of Performance :

Date of Submission :

Aim: Implementation of Mesh and Nodal Analysis.

Objective: To know mesh analysis and nodal analysis practical. To evaluate of DC circuit

analysis.

Hardware / Software / Apparatus used:

Sr.No. Equipments and instruments

required

Specifications Quantity

1 Resistance 220 3

2 Ammeter 0 – 10 mA 3

3 Regulated power supply 0-32V, 0-2A 1

4 Breadboard Standard 1

5 Multimeter Digital 1

6 Connecting Wires -------- 10

Theory: 1.Mesh Analysis

In this method KVL (Kirchhoff’s voltage law) is applied to each mesh in terms of mesh currents.

Each mesh is assigned a separate mesh current and KVL is applied to write equations in terms of

unknown mesh currents.

Kirchhoff’s Voltage Law

According to this law, in any closed loop the algebraic sum of voltage drop across each element

and emf in the loop is equal to zero.

In the network shown,

ABEFA - Mesh 1, BCDEB – Mesh 2.

E1 and E2 - Emfs

I1 and I2 - Mesh Currents

According to Mesh Analysis -

Applying KVL to mesh 1 –

I1 R1 – (I1-I2) R3 +E1 = 0

Applying KVL to mesh 2 –

I2R2 – (I2-I1) R3 –E2 = 0

2. Nodal Analysis

This method is based on KCL (Kirchhoff’s current law) to determine voltages of different nodes

with respect to reference node.

Kirchhoff’s Current Law

According to this law the algebraic sum of all the currents meeting at a node is always equal to

zero. In the given network, let incoming currents are positive and the outgoing currents are

negative. Then by Kirchhoff’s current law,

I1+I2-I3 = 0 i.e. I1+I2 = I3

(Sum of incoming currents = Sum of outgoing currents)

PROCEDURE:

Make the connections as shown in the diagram

1. Apply E1 and E2.

2. Measure the voltages across R1, R2 and R3 (KVL).

3. Measure the currents through R1, R2 and R3 (KCL).

4. Take at least five reading by varying the applied voltages E1 and E2.

OBSERVATION:

1) For Mesh Analysis

Sr no. E1

(V)

E2

(V)

V1

(V)

V2

(V)

V3

(V)

1

2

3

4

5

2) For Nodal Analysis

Sr no. E1

(V)

E2

(V)

It=I3

(mA)

I1

(mA)

I2

(mA)

1

2

3

4

5

CONCLUSION: Thus we have calculated currents and drop across resistors in an electrical

circuit using mesh and nodal analysis.

Signature & Remarks:

EXPERIMENT NUMBER: 02

Date of Performance :Date of Submission :

Aim: Verification of Superposition Theorem

Objective: To evaluate fundamental of circuit analysis using superposition theorem.


Sr.No. Equipments and instruments required Specifications

Quantity

1 Resistances 220 Ω 3

2 Ammeters 0-10mA 3

3 Regulated power supply 0-32V,0-2A 1




Theory:Superposition Theorem Statement:-In any linear bilateral network, containing more than one energy source, the resultant currentflowing through any resistance is equal to the algebraic sum of all the currents flowing throughthe same resistance when each energy source acts alone while all other sources are replaced bytheir internal resistances(if any).

When the batteries E1 and E2 both are acting, then let I1, I2 and I3 are the currents flowingthrough the resistances R1, R2 and R3 respectively. These are resultant currents in the abovenetwork when E1 and E2 both are in network.Case 1:When source E1 is acting alone and E2 is replaced by short circuit.I1’, I2’ and I3’ are currents flowing through R1, R2, R3 respectively. In this case, currentsI1’and I2’ are in () direction while I3’s in () direction.Case2:When source E2 is acting alone and E1 is replaced by short circuit.I1’, I2’ and I3’ are currents flowing through R1, R2, R3 respectively. In this case, currents I1’and I2’ are in () direction while I3’s in () direction. Resultant currents I1, I2 and I3 flowing through R1, R2 and R3 are given byI1= (I1’)-(I1’’), I2= (I2’’)-(I2’), I3= (I3’) + (I3’’)

Circuit diagram:

1. Considering both sources (E1 and E2) –

2. Considering E1 source acting alone –

3. Considering E2 source acting alone

PROCEDURE:

1. Make the connections as shown in the circuit diagram.2. Apply E1,E2 and measure the currents through R1,R2 and R3 (i.e.I1, I2 and I3)3. Take at least five readings by varying the applied voltage E1 and E2

4. Remove the battery E2 from the circuit. Now E1 source is acting alone in the circuit5. Measure the currents through R1,R2 and R3 (i.e.I1’, I2’ and I3’)6. Take at east five readings by varying the applied voltage (i.e.E1).7. Remove the battery E1 from the circuit. Now E2 source is acting alone in the circuit8. Measure the currents through R1,R2 and R3 (i.e.I1’’, I2’’ and I3’’)9. Take at least five readings by varying the applied voltage (i.e.E2).10. Verify the theorem.

OBSERVATION1. Considering both sources (E1 and E2)

Sr No. E1

(V)E2

(V)It=I3

(mA)I1

(mA)I2

(mA)

1

2

3

4

5


Sr No. E1(V)

It=I3’(mA)

I1’(mA)

I2’(mA)

1

2

3

4

5


Result:I1= (I1’)-(I1’’)I2= (I2’’)-(I2’)I3= (I3’) + (I3’’)

Conclusion:

From above experiment, superposition theorem is verified.


Sr No. E2

(V)It=I3’’(mA)

I1’’(mA)

I2’’(mA)

1

2

3

4

5



Aim:-Verification of Thevenin Theorem / Maximum Power Transfer Theorem.

Objective:To evaluate D circuit analysis using Thevenin theorem & to concept of maximum power transfer theorem in electrical engineering.


Sr.No. Equipments and instruments required Specifications Quantity

1 Resistance 220Ω

1.2kΩ

4

4

2 Ammeter 0-20mA 1

3 Regulated power supply 0-32V,0-2A 1



6 Connecting Wire -------- 5

Theory: Maximum Power Transfer Theorem states that “maximum amount of power will betransferred to the load resistance when load resistance value is equal to Thevenin’s or Norton’sresistance of the network supplying the power” i.e. RL = RTH or RL = RN

The maximum power is given by PLMAX = VTH

2 / (4 RTH) Where,

VTH = Thevenin’s equivalent voltage or open circuit voltage across terminals A and B

RTH = the equivalent resistance of the network between terminal A and B, when all sources are

replaced by their internal resistance values.

Procedure:-1. Make the connections as shown in the diagram.2. Remove load resistance (RL) from the given network3. Apply voltage E4. Find the Vth which is the open circuit voltage across terminal AB(across R2)5. Find IL

6. Take at least five readings for Vth and IL by varying V

Observation:

SrNo.

E(V)

RL IL

(mA)

(Practically)

PL = IL2 RL

(mW)

(Practically)1

5 V2345

Conclusion : From above experiment, both Thevenin’s theorem and maximum power transfer theorems are verified.


SrNo.

E(V)

RL Vth

(V)(Practically)

IL

(mA)

(Practically)1 4 V

1.2K2 6 V3 8V4 10 V5 12 V



Aim: To study series RLC series single phase AC circuit

Objective: To impart knowledge of fundamental of single phase AC circuit & evaluate its various parameter


Sr.

No.

Equipments and instruments required

Specifications Quantity

1 Rheostat (Resistance) 5A,100Ω 1

2 Inductor (Choke) Xmer Choke 1

3 Capacitor 470 µf 1

4 AC Ammeter 0-5A, AC 1

5 AC Voltmeter 0-600V, AC 1

6 AC Wattmeter 5/10A,150/60 Ω

1


Theory: In series R-L-C circuit when AC voltage is applied, current either leads or lags voltage and itdepends upon the XL and XC values as in the cases mentioned below.

CASE: 1 (XL > XC): The circuit becomes inductive; current ‘I’ lags the voltage ‘V’

CASE: 2 (XC > XL): The circuit becomes capacitive; current ‘I’ leads the voltage ‘V’

CASE: 3 (XC = XL): The circuit behaves neither inductive nor capacitive, it is resistive in nature. This is also known as “Series Resonance”. Current ‘I’ is in phase with the voltage ‘V’.

V=VR+VL+VC

VR = I * R, VR and I are in phase

VL = I * XL VL leads current by п/2

VC = I* XC VC lags current by п/2

Circuit Diagram

Procedure:

1. Connect the circuit as shown in figure.2. Apply some voltage to circuit.3. Note down the current and voltage across R,L and C.4. Repeat the step (3) by increasing the applied voltage.5. Take at least five readings.6. Draw vector diagram for any two sample reading

Observations:

Sr. No. VT IT WP VR VL VC

Result:The values of Inductor Capacitor and resistors are ………..….

Conclusion / Outcome:

Thus we have studied series RLC circuit.




Aim: To verify relation between in three phase AC circuit

1. Line voltage and Phase voltage2. Line current and Phase current

Objective: To evaluate fundamental of three phase AC circuit.


Sr. No. Equipments and instruments required Specifications Quantity

1 Dimmer stat (Auto Transformer) 3Ф,440v,50Hz 1

2 Rheostat 5A,100Ω 3

3 AC Wattmeter 5/10A,150/60 Ω 2



6 Connecting wires ----- 10

Theory: In three phase system, balanced, star connected system, the current flowing through the neutral wire is the phasor sum of all the line currents/ phasor currents and it is always zero.

(in balanced system, neutral current is not equal to zero.)

i.e.IN = IR + IY + IB =0

In a balanced star connected system, the voltage between any two lines or phases is known as theline voltage and the voltage between any one phase and the neutral wire is called the phase voltage. In balanced system all the line voltages are same and all the phase voltages are same. Also line voltage=√3 VPH

In the balanced star connected system, the voltage between any lines is known as the line currentand current flowing through any phase current. All the line currents are the same and all phasecurrents are same. Also the line current=phase current i.e. IL=IPH

Circuit Diagram:

Procedure:

1. Construct the star connected balanced load2. Set the Dimmerstat at its zero position and switch ON the three phase supply3. Set the value of line voltage by varying the Dimmerstat4. Note down the readings of VL,IL,VPH and IPH

Result and / or Observations:

1. Line Voltage and Phase voltage=

2. Line Current and Phase Current=





Aim: To measure the power of three phase load by two wattmeter method.

Objective: To illustrate the method of measurement of power in three phase AC circuit.


Sr. No. Equipments and instruments required Specifications Quantity

1 Dimmer stat (Auto Transformer) 3Ф,440v,50Hz 1

2 Rheostat 5A,100Ω 3

3 AC Wattmeter 5/10A,150/60 Ω 2



6 Connecting wires ----- --

Theory:

Wattmeter is a device which senses the current, voltage and angle between voltage and current(i.e. power factor) and gives the power reading directly in Watts. Means the wattmeter is used tomeasure the power in AC circuits

It consists of two coils to sense current and voltage. The coil sensing current is called currentcoil, it is always connected in series with the load hence constructed with thick conductor andless number of turns. The other coil is called voltage coil also called as pressure coil, this isalways connected in parallel with the load so wound with thin wire and large number of turns. It

is important to note that the wattmeter senses the angle between current phasor and voltagephasor sensed by respective coils. So

W = Vpc * Ic * Cos (Vpc^Ic)

Where Vpc is the voltage across pressure coil and Ic is current through current coil. Angle is decided from phasor diagram.

Two Wattmeter Method:

The current coils of two wattcmeters are connected in any two lines while the voltage coil of each wattmeter is connected between its own current coil terminal and a line without current coil as shown in the figure. The connections are same for both star and delta connected loads. The total power dissipated in the three phase circuit is equal to algebraic sum of two wattmeter readings. If W1 and W2 are two wattmeter readings then total power is

W= W1 + W2 (Watts)

Circuit Diagram:

Procedure:

1. Make the connections as shown in the circuit diagram2. Apply source voltage to the circuit 3. Make all the line current same by adjusting the load4. Take the corresponding readings5. Repeat steps 2, 3 and 4 for more readings6. Verify the relations W1+W2= √(3) * VL * IL

Observations:

Sr No VL Volts

IL Ampere

W1

Watts

W2

Watts

W1+W2

Watts

P=√ (3)*VL*IL

1

2

3

4

5

Result :

Power consumed by 3 phase system is obtained by simply adding two wattmeter readings.


From the above experiment it has been verified that the power consumed by 3 Phase load can be measured by two wattmeter method.




Aim: To study O.C and S.C. Test on single phase transformer.

Objective: To impart knowledge of basic operation and to evaluate performance of single phase transformer


Sr.No. Equipments and Instruments Required Specifications Quantity

1 Transformer 230/18V,50Hz 1

2 Dimmer Stat (Auto Transformer) 3 phase,440V 1

3 AC Voltmeter 0-300V 1

4 AC Ammeter 0-5A 1

5 AC Wattmeter 0-150V,2A 1

6 Multimeter ------------- 1

7 Connecting Wires 1/18 SWG 10

Theory:

The parameters of the equivalent circuit of a transformer can be obtained by using the open circuit (O.C.) and the short circuit (S.C.) test on the transformer.Open Circuit Test (O.C. Test)The setup for O.C. test of a transformer is shown in fig.

The primary winding of transformer is connected to the rated ac voltage by means of

using a variac. A voltmeter is connected across the primary winding to measure the primary voltage

An ammeter is used for measuring the primary current and the wattmeter is connected to

measure the input power. The secondary is open circuited because it is open circuit(O.C.) test. Sometimes a

voltmeter is connected across the secondary to measure V2=E2 Note that the AC supply voltage is applied generally to the low voltage side and the

higher voltage side is used as secondary

Short Circuit Test (S.C.Test):-

The setup for carrying out the S.C. test on transformer is shown in fig. Variac is used to adjust the input voltage precisely to the rated voltage The voltmeter is connected to measure the primary voltage. The ammeter measure

the short circuit rated current Isc and the wattmeter measure the short circuit inputpower

The secondary is short circuited with the help of thick copper wire Generally the high voltage side is connected to the AC supply and the low voltage

high current side is shorted.

Circuit Diagram:

S in g le p h a s eA C s u p p ly

M

C

L

V

W A T T M E T E R

V A R I A C

L O W V O L TA G E S ID E H I G H V O L TA G E S ID E

O P E N C I R C U I T T E S T

S in g le p h a s eA C s u p p ly

M

C

L

V

W A T T M E T E R

V A R I A C

L O W V O L TA G E S I D E H IG H V O L TA G E S ID E

S H O R T C I R C U I T T E S T

Observations:

For open circuit test:-

Sr. No. VOC IOC W1 V2

1

For short circuit test:-

Sr. No. VSC ISC WSC I2

1

Calculations :


WOC = (IOC) (VOC) COS Ø0

Iw = (IOC) COS Ø0

Im = (IOC) SIN Ø0

Ro = voc / Iw

Xo = Voc / Im


WSC = (ISC2 ) R01

Z01 = VSC / ISC

X01= √(zo1)2 – (R01)2

Result:


Value of Ro=

Value of Xo=

For short circuit test:-

Value of Ro1=

Value of Xo1=





Aim: To study series RLC circuit.

Objective: To impart knowledge of basic function of RLC resonance circuit


Sr.No.

Equipments and Instruments Required

Specifications

Quantity

1 Resister 100 Ω 1

2 Inductor 100 mH 1

3 Capacitor 0.1 µF 1

4 Breadboard 1

5 Function Generator 1

6 Ammeter 0-100 A 1

Theory: In series RLC circuit current lags behind or leads voltage depending upon values of XL and XC. XL

causes the total current to lag behind the applied voltage while XC causes total current to lead the appliedvoltage. When XL >XC the circuit is predominantly inductive , and when XC>XL circuit is predominantlycapacitive.

In series RLC circuit resonance may be produced by varying the frequency.At a certain supply frequency, called resonance frequency (fr), the inductive reactance becomes equal tocapacitive reactance and the net reactance(X) becomes zero. Therefore Impedance (Z) of the circuit becomes purely resistive (i.e Z =R). In other words the whole circuitbehaves as a purely resistive circuit and the current remains in the phase with the applied voltage (pf=1).i.e. The applied voltage and the circuit current are in phase. If such condition occurs in the seriescircuit, it is termed as series resonance

Circuit Diagram:

Procedure: 1. Connect the circuit as shown in figure. 2. Apply a fixed voltage (5v) through function generator to the circuit.

3.The frequency of the circuit is varied in steps and note down the corresponding ammeter readings. 4. At resonance frequency current becomes maximum. 5.Compare practical value of resonance frequency with theoretical value.

fr =1

2 π √ LC .

Observation:

Sr. No.

Frequency

Voltage (VR)

Current

(V R

R )

1.

2.

3.

Result :Resonant frequency by calculation - _________ Hz.Resonant Frequency by observation - __________Hz.Conclusion: In series resonance circuit current becomes maximum at resonant frequency.


Documents

EXPERIMENT NUMBER:01 Date of Performance : Date of ...dmce.ac.in/newdmcewebsite/others/resource/Humanities/BEE MANUAL18-19.pdf · PROCEDURE: Make the connections as shown in the diagram