22 Ec Lab Master Manual 14

8/11/2019 22 Ec Lab Master Manual 14

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Electrical Circuits Lab Manual Page 1 of 73

Department ofElectrical Electronics Engineering Swarnandhra college of Engineering Technology

OBJECTIVE

This lab is very important to the EEE engineering students, to know the practical

behavior of both passive and active sources elements. The students are direct interact with

the active and passive elements / sources and the students can gain the knowledge about

the voltage drops, branch currents and power calculations. For different circuits likes RLC

circuits, theorems, resonance, open circuit and short circuit parameters with both the

circuits (AC & DC). After completion of the lab the students are in a position to do the

design of a given circuits and calculations like voltage, currents and they can prove the

practical and theoretical results are same. This lab also useful for the students of EEE, ECE,

MECH & CIVIL.

ELEMENT NAME SYMBOL

RESISTOR

VARIABLE RESISTOR

INDUCTOR

CAPACITOR

VARIABLE CAPACITOR

DC SOURCE

AC SOURCE

AMMETER A


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Department ofElectrical Electronics Engineering Swarnandhra college of Engineering Technolog


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Expt. No: 01

Date:EXPERIMENTAL DETERMINATION OF THEVENINS AND

NORTONS THEOREMSAIM:

To determine Thevenins & Nortons equivalent circuit and verify thesame theorems for the given circuit where V 1 = 15 Volts; V 2 = 10 Volts;R1=1K ; R 2 = 2 K ; R 3 = 3 K .APPARATUS REQUIRED:

THEORY:

THEVENINS THEOREM STATEMENT:

Thevenins theorem states that any twoterminal linear network having number of voltage,current sources and resistance can be replaced by asimple equivalent circuit consisting of a single voltagesource in series with a resistance ,where the value ofthe voltage source is equal to the open circuit voltageacross the two terminals of the network, andresistance is equal to the equivalent resistancemeasured between the terminals with all the energysources are replaced by their internal resistances.

CIRCUIT DIAGRAM: THEVENINS EQUIVALENT CIRCUIT:

+

-

Sl.No.

Particulars Range &Type Quantity

1 Circuit Board - 1

2 Ammeter 0-200 mA, MC 1

3 Voltmeter 0 30 V,MC 1

4 Multi Meter digital 1

5 Regulated Power Supply 0 - 30 V 1

6 Patch Cords - 10


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STEP BY STEP PROCEDURE:STEP I:

To find the current through the load resistance R L divide the network

in to two parts A and B by separating the R L from the original network.

STEP II: FIND VO.C.

VO.C. = Vth = 1

15 R

V + 2

10 R

V = 0

= 1000

15V +

200010V

= 0

= V

20001

10001 =

200010

100015

= 3 V = 40 V = 3

40 = 13.33 V Vth = 13.33 V

STEP III: FIND RTh.

Replace all the current sources, voltage sources with open circuit and short circuit respectively.

Rth = 21

21

R R R R

= 2121 X

= 0.667 K

STEP IV: THEVENINS EQUIVALENT CIRCUIT:

Now, we can replace the network in to Thevenins equivalent circuit as shown in

figure below.


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IL = Lth

th

R R

V =

3103667.033.13

X = 3.64 X 10 3 A

Therefore Vth = 13.33 V

Rth = 0.667 K IL = 3.64 m A

PROCEDURE: 1. Connections are made as per the circuit diagram. 2. Measure the open circuit voltage across the load terminals. 3. Measure the equivalent resistance (Rth) of the circuit looking form the load

terminals. 4. From the equivalent circuit diagram, calculate the value of load current.

Sl. No. DESCRIPTION THEORITICAL PRACTICAL

1. V Th (VO.C.) 13.33 V

2. R Th 0.667 K

3. I L 3.64 m A

RESULT:

Thevenins

Theorem

is

verified

since

the

practical

values

are

same

with

the

theoretical values.


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NORTONS THEOREM STATEMENT:

Nortons theorem states that any twoterminal linear network having number of voltage,current sources and resistance can be replaced bya simple equivalent circuit consisting of a singlecurrent source in parallel with a resistance ,wherethe value of the current source is equal to thecurrent flowing through the short circuit betweenthe two terminals of the network, and resistanceis equal to the equivalent resistance measuredbetween the terminals with all the energy sourcesare replaced by their internal resistances.

CIRCUIT DIAGRAM: NORTONS EQUIVALENT CIRCUIT:

STEP I:THEORITICAL CALCULATIONS FOR IN:

Divide the network in to two parts A and B.

STEP II:Calculations for IN (Isc): Apply KVL to loop ABEFA

1000 I1 15 = 0 I1 = 1000

15 = 15X10 3 A

Apply KVL to loop BCDEB

2000 I2 + 10 = 0 I2 = 2000

10 = 5X103 A

ISC = IN = I1 I2 = 15X10 3 ( 5X10 3) = 20 m A


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STEP III: CALCULATIONS FOR RN:

Replace the current sources and voltage sources with its equivalents and find the Req at the terminals BE

Req = RN = 21

21

R R R R

= 33

33

102101102101

X X X X X

= 0.667 K

RN = 667 STEP III: CALCULATIONS FOR IL:

Calculation for IL: connect the current source IN< Nortons resistance RN, and load resistance RL in parallel

IL = IN X L N

N

R R

R = 20 X 10 3 x 3103667

667 X

= 3.64 X 10 3 A IL = 3.64 m A

IN = 20 m A; RN = 667 IL = 3.64 m A TABULAR FORM:

Sl. No. DESCRIPTION THEORITICAL PRACTICAL

1.

2.

3.

ISC (IN)

RN

IL

PROCEDURE: 1. Connections are made as per the circuit diagram. 2. Measure the short circuit current flows through the load terminals. (ISC) 3. Measure the equivalent resistance of the circuit looking form the load terminals.

4. From

the

Nortons

equivalent

circuit

diagram

and

from

the

measured

values,

calculate the current flows through the load.

RESULT: Nortons theorem is verified since the practical values are equal to the theoretical

values.


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Viva Voce Questions:

(a)

For Thevinins Theorem:

1. Statement of Thevinins theorem.2. Applications of Thevinins theorem.3. Define electric circuit.4. Define electric network.5. Define energy.

6. Define electric current.7. Define resistance.8. Define inductance.9. Define capacitance.

10. Five coulombs of charge flow past a given point in a wire in 2 sec.how many amps of current is flowing?

(b) For Nortons Theorem:

1. Statement of Nortons Theorem.2. Applications of Nortons Theorem.3. Define EMF.4. Define MMF.5. Define potential difference.6. Define current density.7. Define specific resistance.8. Define conductivity of a material.9. What is positive temperature coefficient?

10. What is negative temperature coefficient?


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Expt. No: 02

Date:

VERIFICATION OF SUPER POSITION THEOREM AND

MAXIMUM POWER TRANSFER THEOREM Aim : To verify superposition theorem and maximum power transfertheorem with resistive loads under DC excitation for the given circuit whereV1 = 5 Volts; V 2 = 5 Volts; R1= 2 K ; R2= 3 k ; R3 = 1 k Apparatus Required:

Sl. No. Particulars Range Type Quantity

1 Circuit Board WB-102 1

2 Ammeter (0 - 200) mA MC 1

3 Voltmeter (0 - 30) V MC 14 Ammeter (0 - 200) ma MC 1

5 Voltmeter (0 - 30) V MC 1

6 Multi Meter Digital 1

7 Resistances 1K , 2K , 3K 1

8 Regulated Power Supply 0 - 30 V DUAL 1

9 Signal generator 1

10 Patch Cords 6CIRCUIT DIAGRAM:

STATEMENT:

The Superposition theorem states that in any linear bilateral network containing

two or more sources, the response in any element is equal to the algebraic sum of the

responses caused by individual sources acting alone, while the other sources are non

operative; that is, while considering the effects of individual sources, other ideal voltage

sources and ideal current sources in the network are replaced by short circuit and open

circuit across their terminals. This theorem is valid only for linear systems. Power cannot be

determined by superposition, since the relation between the power and current or voltage

are nonlinear.


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THEORY:

The principle of Superposition is a combination of additivity property and homogeneity

property. The property of additivity says that the response in a circuit due to no. of sources

is given by the sum of the response due to individual sources acting alone. The property of

Homogeneity says that if all the sources multiplied by a constant then the response is also multiplied by the same constant.

The responses that can be determined by Superposition theorem are listed below:

1. Current in resistance, inductance and capacitance. 2. Voltage across resistance, inductance and capacitance. 3. Current delivered by the independent voltage sources. 4. Voltage across independent current sources. 5. Voltage and current of linear dependent sources. While calculating the response due to individual sources all other sources are made

inactive. Therefore, while calculating the response due to one source, all other voltage

sources and current sources are replaced by short circuit and open circuit respectively

THEORITICAL CALCULATIONS: Apply KCL to node 1

25V

+ 1V

+ = 0

31

121

V = 25 +

35

6263

V = 6

1015

1125

V = 2.273 V

31 1025

X V

I

= 31025273.2

X = 1.364 m A 32 103

5 X V

I

= 31035273.2

X = 0.909 m A

33 R

V I = 3101

273.2 X

= 2.273 m A

Negative sign represents the current directions are opposite to the conventional

current direction and the current values are the final values when both the voltage sources

connected in the circuit.

The current values in each branch when the individual voltage source acting alone

and other voltage source replaced by the current source is given below

STEP I: V1 SOURCE ACTING ALONE: When the voltage source V1 acting alone and the other sources V1 replaced by the zero internal resistance; the current values in different branches are given below.

Req = R1 + R2 | | R3 = 2 + 13

13 X = 2.75 K .

35V


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I = I1| = eq R

V 1 = 31075.25 X

= 1.82 m A.

I2| = 32

3|

1

R R

XR I = 1.82 X 10 3 X 3

3

10)31(101 X

X = 0.455 m A.

I3| = 32

2|

1

R R

XR I = 1.82 X 10 3 X 3

3

10)31(103 X

X = 1.365 m A.

STEP II:

V2 SOURCE ACTING ALONE: When the voltage source V2 acting

alone and the other sources V1 replaced by the zero internal resistance; the current values in different branches are given below.

Req = R2 + R1 | | R3 = 3 + 12

12 X

= 3.667 K .

I = I2|| = eq R

V 2 = 310667.35 X

= 1.364 m A.

I1|| = 32

3||

2

R R

XR I = 1.364 X 10 3 X 3

3

10)21(101 X

X = 0.455 m A.

I3|| = 32

1

||

2

R R XR I = 1.364 X 10 3 X 3

3

10)21( 102 X X = 0.909 m A.

I1 = I1| I1|| = 1.82 0.455 = 1.364 m A

I2 = I2|| I2| = 1.364 0.455 = 0.909 m A

I3 = I3| + I3|| = 1.365 + 0.909 = 2.274 m A

By the observation the algebraic sum of the responses are equal to the responses

when both the voltage sources acting simultaneously. Hence the superposition theorem is

verified.

PROCEDURE:

1. Make the connections as per the circuit diagram while the two voltage sources acting simultaneously.

2. Note down the values of currents in each branch. 3. Replace one voltage source by the internal resistance and note down the current

values in each branch while the other voltage source acting alone. 4. Repeat the same while placing V2 in the circuit and the other source V1 replaced by

the short circuit.


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TABULAR FORM: When both the voltage sources acting in the circuit.

DESCRIPTION I1(mA) I2(mA) I3(mA)

Theoretical

Practical

When the voltage source V1 is acting alone

DESCRIPTION I11(mA) I12(mA) I13(mA)

Theoretical

Practical

When the voltage source V2 acting alone

DESCRIPTION I11 1(mA) I11 2(mA) I11 3(mA)

Theoretical

Practical


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MAXIMUM POWER TRANSFER THEOREM AIM: To verify Maximum power transfer theorem on D.C and on A.C with resistive and

reactive loads.

APPARATUS:

Sl. No

DESCRIPTION SPECIFICATION QUANTITY

1. Transistorized power supply (T.P.S) (030)V 1

2. Signal generator 13. Decade Resistance Box (D.R.B) 3

4. Decade Inductance Box (D.I.B.) 15. Decade Capacitance Box (D.C.B.) 1

6. Multi meter Digital 17. Bread Board 1

8. Connecting probes L.S.

CIRCUIT DIARAM:

STATEMENT FOR RESISTIVE LOAD:The maximum power transfer theorem states that the maximum

power is delivered from a source to its load when the load resistance isequal to the source resistance.EXPLANATION:

A variable resistance RL is connected to a D.C. source network as

shown in figure 5.2 (a) while figure 5.2 (b) represents the Thevenin voltageV0 and Thevenin resistance Rth of the source network.


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The main aim is to determine the value of RL such that it receives maximum power

from the D.C. source.

With reference to the figure I = )(

0

Lth R R

V

While the power delivered to the resistive load is

PL = I2 RL = X R R

V

Lth

2

0

RL

PL can be maximized by varying RL and hence, maximum power can be delivered

when 0 L

L

dR

dP

However, L

L

dR

dP=

22020222

1 Lth

L L L

L Lth

Lth

R RdR

d RV RV

dRd

R R R R

= 4

1

Lth R R Lth L Lth R R X RV V R R 2

20

20

2

=

42

0 2

Lth

L Lth Lth

R R

R R R R RV =

3

20

Lth

Lth

R R

R RV

Finally

32

0

Lth

Lth

R R

R RV = 0

which gives Rth RL = 0 (or) Rth = RL

Hence, it has been proved that power transfer from a D.C. source network to a resistive network is maximum when the internal resistance of the D.C. source network is

equal to the load resistance.

Pmax = 2

20

Lth

L

R R

RV =

22

0

2 th

th

R

RV =

th R

V

4

20

This is the power consumed by the load.

THEORETICAL CALCULATIONS: STEP 1:

Remove the load resistance and find Thevenins Resistance (RTh) of the source network looking through open circuited load terminals.

RTh = 3131 X

+ 2 = 2.75 K

STEP 2: As per the maximum power transfer theorem, this RTh is the load resistance of the

network i.e. RL = RTh that allows maximum power transfer


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RL = RTh = 2.75 K

STEP 3: Find the Thevenins voltage V0

across the open circuited terminals

Vo.c. = VAB = Vth = I3 R Req = 1+3 = 4 K

I = eq R

V =

310410

X = 2.25 X 10 3 A

I3 = 2.5 m A VTh = 2.5 X 10 3 X 3 X 10 3 = 7.5 V

STEP 4:

Maximum power transfer is given by th R

V

4

20

Pmax = 32

1075.245.7

X X = 5.11X10 3 W

RL < Rth = 2 K

P = 2

20

Lth

L

R R

RV =

2332

10275.2

1025.7 X X = 4.99 X 10 3 W = 4.99 m W

RL > Rth = 3.5 K

P = 22

0

Lth

L

R R

RV = 23

32

105.375.2

105.35.7 X X = 5.04 X 10

3

W = 5.04 m W

PROCEDURE for D.C. & A.C.:

1. Connections are given as per the circuit diagram shown in figure.52. Adjust the RPS 1 and RPS 2 for the value of Voltage sources V 1 &V 2 given

in the problem and note down the voltmeter reading3. Connections are given as per the circuit diagram shown in figure.64. Measure the resistance value by multimeter and note down the reading5. Connections are given as per the circuit diagram shown in figure.7

6.

Adjust R L = R s for getting maximum power, set the RPS 1 for the valueof V S that has found out in step 2, and note down the ammeter readingand find the power.

7. Now adjust R L < R s , note down the ammeter reading and find thepower.

8. Now adjust R L > R s , again note down the ammeter reading and find thepower.

9. Draw the graph between Power and load current and verify where themaximum power is obtaining in the graph.


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MODEL GRAPH.

MAXIMUM POWER TRANSFER THEOREM for A.C. CIRCUITS: CIRCUIT DIAGRAM:

STATEMENT: A load impedance being connected to a linear network receives maximum power,

when the load impedance is equal to the complex conjugate of the internal impedance of the active network as seen from the load terminals.

EXPLANATION:

The active linear network to which the load is connected is Thevenised in to a single voltage source Voc in series with an internal impedance ZTh with respect to load terminals as shown in above figure.


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Then the power transferred to the load is PL = IL2 RL

PL = 222

20

Lth Lth

LC

X X R R

RV = 0

Assuming that

only

XL

is

variable

and

RL

is

fixed

for

maximum

power

L

L

dX

dP =

0

L

L

dX dP

= 222

2 2

Lth Lth

Lth LOC

X X R R

X X RV = 0

Hence XL = Xth (2)

Now to maximize PL further, let RL also be varied so that L

L

dR

dP = 0

=

222222 2

Lth Lth

Lth L Lth LthOC

X X R R

R R R X X R RV

i.e. = 2 Lth R R 22 Lth Lth L X X R R R = 0 (3) Substituting the equation (2) in equation (3)

Lth Lth R R R R = 0 Rth 2 + RL2 = 0 RL = Rth Therefore for the power transferred to the load is maximum in which RL and XL are both variable ZL = Rth j Xth conjugate of Zth

th L Z Z

The maximum power transferred to the load under this condition is Pmax = 2

2

2 th

LOC

R

RV

Pmax = th

OC

R

V

4

2

(RL = Rth )

This is independent of Xth

THEORITICAL CALCULATIONS:

To find Zth open the terminals of load and determine the resistance looking form the terminals.


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R = 1000

XC = fC 2

1 = 610475021

X X X = 67.73

Zth = 22 C X R

Zth = 22 63.671000 = 1002.3 Vth is the open circuit voltage = 5 V

Pmax = 2

2

2 th

thOC

R

RV =

th

OC

R

V

4

2

= 100045 2

X = 6.25 m W

TABUALR FORM for D.C. INPUT: Rth = 2.75 K , Pmax = 5.11 m W

Sl. NO. RL ( ) Voc (V) POWER (mW)

For A.C. INPUT:

Rth = 1 K , Pmax = 6.25 m W

Sl. NO. RL ( ) Voc (V) POWER (mW)

RESULT: Maximum power transfer theorem is verified.


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Viva voce Questions:

1. Statement of Superposition theorem.2. Applications of Superposition theorem.3. Define RMS value of Complex wave.4. What is the expression for RMS value of complex wave?5. What is meant by Harmonic components?6. What is a linear element? give few examples.7. What is a Nonlinear element? give few examples.8. What is an ideal voltage source?9. What is an ideal current source?10. Super position theorem is valid for ______


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Expt. No: 03

Date:VERIFICATION OF COMPENSATION THEOREM

AIM: To verify compensation theorem.

APPARATUS REQUIRED:

Sl. No. Apparatus required Specification Quantity

1. T.P.S (0-30) V 1

2. Ammeter (0-200) mA 1

3. Resistor 1 1

4. Resistor 2.2 k 1

5. Resistor 500 1

STATEMENT:It states that any element in the linear bidirectional network, may be

replaced by a voltage source of magnitude equal to the current passingthrough the element multiplied by the value of the element, provided thecurrents and voltages in other parts of the circuit remains unaltered .THEORY:STATEMENT:

If the impedance Z of a branch in a network in which a current I flows is changed by a finite amount Z, then the change in the currents inall other branches of the network may be calculated by inserting a voltagesource of -I Z into that branch with all other voltage sources replaced bytheir internal impedances.

This theorem is useful in finding the changes in current or voltagewhen the value of resistance is changed in the circuit. Consider the networkcontaining a resistance R shown in figure 1. A small change in resistance Ras shown in figure 2, causes a change in current in all branches. This

current increment in other branches is equal to the current produced bythe voltage source of voltage I R which is placed in series with alteredresistance as shown in figure 3.

I

R N.W

Fi 1

R

I

R

N.W

Fi 2

I R

R

R

N.W

Fi 3


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Observations:

Sl. No. Description Theoretical Practical

1

2

3

Fig. 4

Fig. 5

Fig. 6

I1 = 2mA

I2 = 0.421mA

I3 = 1.578mA

R+ R=2K

I2

I3=I1 I2

I1 A

Circuit diagrams for Compensation theorem

1K 2K

1K3K10V

(0200) mA+

+

A1K 2K

1K3K

(0200) mA+

2V +

R

R

R)(R

A1K 2K

2K310V

(0200) mA+

+

Figure 4

Figure 5

1K

V=I 1*

R

Figure 6


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Theoretical calculations :

Step1: Find the current in theresistor that will be changed (i.e.,

R):

2000 1000 30001000 2500 2.5

2000 1000 3000eq R K

104

2500T I mA

Applying current divide rule, we get

3

13000

(4 10 ) 2(2000 1000) 3000

I mA

I1 = 2 mA Step2: Find the changing current

in the changed resistor(ie., R+ R)by applying Compensationtheorem:

I2 = 0.421 mA

According to theorem first we remove

existing source and connect acompensating voltage source V=I 1 x R =(2x10 -3 ) x 1000 = 2 V, along with thechanged resistor as shown in figure andfind the current.Now for finding current through R+ R, first

find the total resistance as

1000 30002000 1000 1000 4750 4.7

1000 3000eq R K

22

0.4214700T

I I mA

Step3: Find the changing currentin the changed resistor(ie., R+ R):

I3 = 1.57 mA

2000 2000 30001000 2714 2.71

2000 2000 3000eq R K

103.68

2714T I mA

Applying current divide rule, we get

3

33000(4 10 ) 1.57

(2000 2000) 3000 I mA

Step4: Verify the result:

I3 = I 1-I 2 Therefore theorem is verified.

Step4: Verifying the result:

I3 = I 1 -I 2 = (2x10 -3 ) (0.421x10 -3 )= 1.579 mA = I 3 (obtained)


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PROCEDURE:

1) Connect the circuit diagram as shown in fig4. 2) Give the rated supply to the circuit through TPS and note down the

ammeter reading and say this current be I 1 . 3) Now connect the circuit diagram with the changed resistor and shortcircuiting the voltage source as shown in fig5, giving thecompensation voltage such that 1V I R and note down theammeter reading, say this current be I 2 .

4) Now connect the circuit with the changed resistor as shown in fig6,give the rated voltage and note down the ammeter reading, say thiscurrent be I 3 .

5) Now verify that I 3 = I 1 I 2

RESULT:

Hence Compensation theorem is verified for the given circuit boththeoretically and practically.


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1. Statement of Compensation theorem.2. What are the applications of Compensation theorem?3. What is the usefulness of this theorem?4. What is a unilateral element? Give few examples.5. What is a bilateral element? Give few examples.6. What are the limitations of Compensation theorem?7. What is a lumped element?8. What is a distributed element?9. Define Ohms law & what are its limitations?10. Define resistivity.


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Expt. No: 04

Date:

VERIFICATION OF RECIPROCITY AND MILLMANS THEOREMSAIM:

a) To verify Reciprocity Theorem for the given circuit.

b) To verify Millmans Theorem.

APPARATUS REQUIRED:

Sl.No DESCRIPTION SPECIFICATION QUANTITY

1 T.P.S (0 - 30) V 1

2 Resistance 560 23 Resistance 1k 3

4 Resistance 2k 1

5 Resistance 3k 2

6 Resistance 3.3 k 1

7 Resistance 2.2 k 1

8 Ammeter (0-200) mA 1

9 Bread board WB-102 110 Multimeter digital 1

11 Connecting wires - L.S.

(A) STATEMENT-RECIPROCITY THEOREM:

In any linear bilateral network, if a single voltage source V a inbranch a produces a current I b in branch b ,then if voltage source V a is removed and inserted in branch b will produce a current I a inbranch a. The ratio of excitation to response is same for the twoconditions mentioned above

Figure 5

NW NW


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CIRCUIT DIAGRAM

Consider the network shown in figure below. AA 1 denotes the inputterminals and BB 1 denotes output terminals. The application voltage Vacross AA 1 produces current I at BB 1 . Now if the position of the source ischange from AA 1 to BB 1 the response will be at AA1. According toReciprocity theorem the ratio of input to response should be same in boththe cases

When the excitation is given at the terminals AA 1 the response at BB 1 is given below

R6 = R3 | | R5 = 10005601000560 X

= 358.97

R7 = R2 + R6 = 3000+358.97 = 3358.97


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R8 = R7 | | R4 = 97.4358100097.3358 X

= 770.59

R9 = Req = R8 + R1 = 770.59 + 560 = 1.33 X

103

I = eq R

V = I1 = 31033.1

10 X

= 7.52 m A

I2 = I X 74

4

R R R

= 7.52 X 10 3 X 1560

101 3 X = 1.73 m A

I3 = I2 X 35

5

R R

R = 1.73X10 3X

1560101 3 X = 1.11 m A

For 10V excitation the response is 1.11mA. If the excitation is changed from AA1 to BB1 terminals the response at AA1 is given below

R6 = R1 | | R4 = 1560

1000560 X = 358.97 ; R7 = R6 + R2 = 3000 + 358.97 =3358.97

R8 = R7 | | R5 = 97.4358 100097.3358 X = 770.59

R9 = Req = R8 + R3 = 770.59 + 560 = 1.33 X 10 3

I = eq R

V = I3 = 31033.1

10 X

= 7.52 m A

I2 = I3 X 25

5

R R

R = 7.52 X 10 3 X

97.4358101 3 X = 1.73 m A

I3

= I2

X

14

4

R R

R

= 1.73X10

3 X

1560

101 3 X =

1.11

m

A

For 10V excitation at BB1 the response is 1.11mA at the terminals AA1. From the calculation, the ratio of the excitation to response is same as in the

previous condition (when excitation is at AA1 and the response is at BB1). Hence the reciprocity theorem is verified. PROCEDURE:

1. Design the circuit and make the connections as per the circuit diagram. 2. Note down the reading of ammeter at the terminals BB1 while giving the excitation

of 10V at the terminals AA1. 3. Note down the ammeter reading by interchanging the terminals of voltage source

and ammeter. 4. Verify the result with theoretically calculated values.


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TABULAR FORM: Sl. No Condition Excitation

Response Theoretical Practical

1.

2.

Excitation at AA1 & Response at BB1

Excitation at BB1 & Response at BB1

10V

10V

1.11mA

1.11mA

PROCEDURE: 1. Design the circuit and make the connections as per the circuit diagram. 2. Note down the reading of ammeter at the terminals BB| while giving the excitation

of 10 V at the terminals AA| 3. Note down the ammeter reading while interchanging the terminals of voltage

source and ammeter 4. Verify the result with theoretically designed values.

RESULT: Reciprocity theorems is verified, since the practical values are approximately equal

to the theoretical values.

MILLMANS THEOREM:STATEMENT:

Millmans theorem states that in any network if the voltage sourcesV1 ,V2 ,V 3.. V n in series with internal resistances R 1 ,R 2 ,R 3 R n respectively, are in parallel, then these sources may be replaced by a singlevoltage source Vm in series with Rm.

Step I:

Consider given circuit:

Step II:

Now remove load resistance, as we want to findcurrent through R 3 , load resistor is R 3. After removing R 3 as shown in fig.11, the circuitlooks as many branches, containing voltagesource and resistance in series, are connectedin parallel, there by we can obtain a singlebranch containing a voltage source andresistance in series by applying Millmanstheorem.

Fig.11

+ +

- -


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Step III: Obtaining Millmans equivalent circuitas shown:

11

1G

R & 2

2

1G

R;

1 2

1m R G G ;

1 1 2 2

1 2m

V G V GV

G G ; Fig.12 Step IV: Now connect the loadresistor ie., R 3 across the Millmansscircuit as shown in Fig.13 and findthe load current as shown:

ml

m l

V I

R R

Verify this current by finding currentin R 3 of the original circuit by someother method.

Fig.13

PRACTICAL CIRCUIT DIAGRAMS:

(0-200) mA

=560

=3K =2.2KR1

R3

A

R2

V2= 10V

Rm

Vm

(0-200)mA

A

R3

Figure 14 (Direct method)

Figure 15 (Millmans equivalent circuit)

V1= 20V


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Observations:

Sl. No.

Theoretical loadcurrent through

Millmansequivalent circuit

Practical loadcurrent through

Millmansequivalent circuit

Practical loadcurrent by

Direct Method

1 7.77 mA 7.8 mA 7.6 mA

Theoretical Calculations : For Millmans Thorem:

Step1: Given Circuit:

whereV1 = 20 Volts;V2 =10 Volts;R1=3K ; R 2=2.2K ; R 3=560

Step2: Remove loadresistance ie., R 3 :

Step2: Removing load resistance ie., R 3 :

As we want to find current through R 3 , thatresistor is the load resistor, therefore afterremoving load resistor circuit obtained isshown in the figure.

Step3: Find Millmansequivalent circuit: 1

1

1 10.333

3000G m

R &

2 2

1 10.454

2200G m

R ;

3 31 21 1

1270.64 1.270.333 10 0.454 10

m R K G G

3

1 1 2 23 3

1 2

20 0.333 10 3 10 0.454 1014.23

0.333 10 0.454 10m

VG VGV V

G G

Vm =14.23V; R m =1.27K ; Step4: Connect load

resistor and find I L :

Connect R 3 across Millmans equivalent circuit

as show in figure and find the load current asbelow:

++

- -


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14.237.77

1270 560m

lm l

V I mA

R R

IL = 7.77 mA

PROCEDURE:MILLMANS THEOREM:

1. Connect the circuit diagram for direct method as shown in figure 14,

give the rated voltage and note down the ammeter reading.

2. For connecting the Millmans equivalent circuit, obtain the Millmans

resistance by1 2

1m R G G where 1 21 2

1 1&G G R R

3. Now obtain the Millmans voltage by1 1 2 2

1 2m

V G V GV

G G

4. Now connect the Millmans equivalent circuit as shown in Fig 15, give

the Millmans voltage through the TPS and note down the ammeter

reading.

5. Verify the currents obtained by direct method and Millmansequivalent circuit and see that both the currents are equal.

RESULT:

Hence Reciprocity theorem is verified both theoretically and

practically. Also Millmans theorem is verified both theoretically and

practically.


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1. Statement of Reciprocity theorem.2. What is the limitation of Reciprocity theorem?3. Statement of Millmans theorem.4. What is the utility of Millmans theorem?5. What are the applications of Reciprocity and Millmans theorems?6. What is meant by active element? Give an example.7. What is meant by passive element? Give an example.8. Determine I using millmans theorem.

9. Define conductance.

10. What is the limitation of reciprocity theorem?

I

3

52

10V 20V


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Expt. No: 05Date:

DETERMINATION OF SELF, MUTUAL INDUCTANCE ANDCOEFFICIENT OF COUPLING

AIM: To determine Self and Mutual inductance of a given transform and alsodetermine the coefficient of coupling.APPARATUS REQUIRED:

Sl. No. APPARATUS SPECIFICATIONS QUANTITY

1 Voltmeter (0-30)V 1

2 Ammeter (0-300)V 1

3 1- Booster Transformer (0-30)V,10A 1

4 1- Variac 230V/(0-270)V,10A 15 1- Transformer 230V/15V 1

6 Connecting wires L.S.

PROCEDURE:CASE 1:

Make the connections as per the circuit diagram shown in fig.1and apply rated voltage to L.V. winding with the help of singlephase meter.

Note down the readings of ammeter, voltmeter readings 1 2V & E

and wattmeter readings 0(W ) in tabular form. Calculate the lone loss current

C 0 0

00

1 0

0 1 0 0

I I cos

Wcos

V I

W V I cos

Magnetizing current 2 2m 0 C(I ) I I and L mE MI L

m

EM ....Henrys

I

1LV

m

VL

2 fI Where 1V is the applied voltage.

f 50HZ CASE 2:

Make the connections as per the circuit diagram shown in fig. 2and apply rated voltage to H.V. winding with the help of variac.

Note down the readings of ammeter, voltmeter, and wattmeter.

Calculate the magnetizing current m(I ) by applying the aboveformula (case 1 step 3).


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Calculate the 2HVm

VL

2 fIH

Now calculate the coefficient of coupling between two coils

HV LV

MK

L L

THEORY:The total flux linkage to current flowing through the circuit is called

self inductance and is given by N

LI

The mutual inductance between the two coils defined as the ratio offlux linkage in first circuit due to current in second circuit to the currentflowing in the second circuit. Thus the inductance 12M is given by

1 21 2 1212

2 1

N NMI I

Their self inductances 1 11 2 221 21 2

N NL ;L

I I

The flux linkage with the circuit 2 due to current in circuit 1 is denoted by

12 .The flux which is actually the part of the total flux produced by

current 1(I ) in the circuit 1.Hence we can write 12 1 11K .similarly for the

flux 21 2 22K .Rewriting the expressions for the mutual inductance

1 21 1 2 2212 12

2 2

2 12 2 1 1121 21

1 1

N N K M M

I I

N N K M M

I I

Let 21 12M M M

2 1 2 1 2 11 2212 21

1 2

N N K K M M M

I I

2 2 11 1 221 2

1 1

N NM K ( )K ( )I I

21 2 1 2M K K L L

Let 1 2K K = K

1 2M K L L

1 2

MK

L L Where K is coefficient of coupling.


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Circuit Diagram:

Circuit diagram for finding L LV & M

Circuit diagram for finding L HV & MTabular Forms:Case 1:

Sl.No.

V1 E 2 Io1 Wo1 Ic1 Im1 L LV M

1 115 230 1.1 44 0.3817 1.031 0.355 0.71

Case 2:Sl.No.

V2 E 1 Io2 Wo2 Ic2 Im2 L HV M

1 230 115 0.55 44 0.191 0.515 1.421 0.71


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Theoretical Calculations:Case 1:

We know that

01 1 01 01

0101

1 01

C1 01 01

W V I cos

W 44cos 0.347

V I 115 1.1

I I cos 1.1 0.347 0.3817A

As2 2 201 1 1c m I I I ,

2 2 2 2m1 01 C1I I I 1.1 0.3817 1.031A

LV

1 1m1

l LV

1LV

m1

V VI

X 2 fL

V 115L 0.355H2 fI 2 50 1.031

Also

M

2 2m1

l

2

m1

E EI

X 2 fM

E 230M 0.71H

2 fI 2 50 1.031

Case 2:

We know that

02 2 02 02

0202

2 02

C 2 02 02

W V I cos

W 44cos 0.347

V I 230 0.55

I I cos 0.55 0.347 0 .191A

As2 2 202 2 2c m I I I ,

2 2 2 2m2 02 C2I I I 0.55 0.191 0.515A

HV

2 2m2

l HV

2HV

m2

V VI

X 2 fL

V 230L 1.421H

2 fI 2 50 0.515

Also

M

1 1m 2

l

1

m 2

E EIX 2 fM

E 115M 0.71H

2 fI 2 50 0.515

Therefore Coefficient of Coupling is,

0.710.9996

0.355 1.421 LV HV

M K

L L

RESULT:Hence for the given transformer we have found the coefficient of

coupling between the primary and secondary coils by conducting suitableexperiments on it.


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1. Define Self Inductance.2. Define Mutual Inductance.3. Define Coefficient of coupling.4. What are the expressions for L, M and K?5. State statically induced and dynamically induced emfs with

examples.6. What is the maximum value for K?7. What is an inductively coupled circuit? Give an example.8. What is dot convention?9. What is meant by time invariant element?10. What is meant by time invariant element?


38/73



Expt. No.: 06Date:

TWO PORT NETWORK (Z & Y PARAMETERS)AIM:

To find the impedance (Z) and admittance (Y) parameters of a givennetwork.APPARATUS REQUIRED:

Sl.No.


1 Transistorized Power Supply (0-30)V 1

2 Resistance 1k 1

3 Resistance 2k 1

4 Resistance 3k 15 Ammeter (0 - 200) mA 1

6 Bread board WB - 102 1



CIRCUIT DIAGRAM:

OPEN CIRCUITS IMPEDANCE PARAMETERS:

To find Z11 , Z21 when I2=0 i.e., open circuited To find Z12, Z22 when I1=0 i.e., open circuit

Y PARAMETERS: Port 2 2 | is Short circuited to find Y11 , Y21 , Port 1 1 | is Short circuited to find Y21, Y22 ,


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THEORY: A two port network is simply a network inside a

black box, and the network has only two pairs of accessible terminals. A port is defined as any pair of terminals into which energy is supplied, or from which energy is withdrawn. Usually one pair represents the input and the other represents the output. Such a building block is very common in electronic systems, transmission and distribution systems.

Figure shows a two port network, or two terminal pair network, in which the four terminals have been paired into ports 11 | and 22 | . The terminals together 11 | together constitute a port. Similarly the terminals 22 | constitute another port. The voltage and current assigned to each of the two ports. The voltage and current at the input terminals are V1 and I1 where as V2 and I2 are specified at the output port. It is also assumed that the currents I1 and I2 are entering into the network at the upper terminals 1 and 2 respectively. The variables of the two port network are V1, V2 and I1, I2. Two of these are dependent variables the other two are independent variables. Z Parameters:

The Z parameters of a two port for the positive directions of voltages and currents may be defined by expressing the port voltages V1 and V2 in terms of the currents I1 and I2. Here V1 and V2 are dependent variables and I1 and I2 are independent variables. The voltage at port 11 | is the response produced by the two currents I1 and I2.

V1 = Z11 I1 + Z12I2 (1) V2 = Z21I1 + Z22I2 (2)

Z11 , Z12, Z21, Z22 are the network functions and are called impedance (Z) parameters. These parameters can be represented by matrices. [V] = [Z] [I]

Thus 2

1

V

V =

22211

1211

Z Z

Z Z

2

1

I

I

The individual Z parameters for a given network can be defined by setting each port current is equal to zero, suppose port 22I is left open circuited then I2 = 0

Z11 = 1

1

I V

at I2 = 0 Z11 is the driving point impedance at port 1 1 | with port 2 2 | open circuited. It is called the open circuit input impedance.

Z21 = 1

2

I V

at I2 = 0 Z21 is the transfer impedance at 1 1 | with port 2 2 | open circuited. It is also called open circuit forward transfer impedance.

Port 11I is left open circuited to make I1 = 0

Z12 = 2

1

I V

at I1 = 0 Z12 is the transfer impedance at port 2 2 | with port 1 1 | open circuited. It is also called the open circuit reverse impedance

Z22 = 2

2

I

V at I1 = 0

Z22 is the driving point impedance at port 2 2 | with port 1 1 | open

circuited. It is also called open circuit output impedance.


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SHORT CIRCUIT ADMITTANCE PARAMETERS: The Y parameters of a two port for the

positive directions of voltages and currents may be defined by expressing the port currents I1 and I2 are dependent variables and V1 and V2 are independent

variables. I1 caused by the effect of two voltages V1 and V2. I1 = Y11 V1 + Y12 V2 and I2 = Y21 V1 + Y22 V2

The individual Y parameters for a given network can be defined by setting each port voltage to zero. If we let V2 be zero by short circuiting port 2 2 | then

Y11 = 1

1

V I

at V2 = 0 Y11 is the driving point admittance at port 1 1 | with port 2 2 | short circuited. It is also called short circuit input admittance

Y21 = 1

2

V I

at V2 = 0 Y21 is the transfer admittance at port 1 1 | with port 2 2 | short circuited. It is also called short circuit forward transfer admittance

Port 11I is short circuited to make V1 = 0

Y12 = 2

1

V I at V1 = 0 Y12

is the transfer admittance at port 2 2 | with port 1 1 I short circuited. It is also called short circuit reverse transfer admittance

Y22 = 2

2

V I

at V1 = 0 Y22 is the driving point admittance at port 2 2 | with port 1 1 | short circuited. It is also called short circuit output admittance.

THEORITICAL CALCULATIONS: Z PARAMETERS:

V1 = Z11 I1 + Z12 I2; V2 = Z21 I1 + Z22 I2 When I2 = 0, i.e. 2 2I is open circuited.

eq RV I 11 = 310)13(

10 X

= 2.5 m A

V2 = I X 1 X 10 3 = 2.5 X 10 3 X 1 X 10 3 = 2.5 V

Z11 = 1

1

I V

at I2 = 0 = 3105.210 X

= 4 K .

Z21 = 1

2

I V

at I2 = 0 = 3105.25.2

X = 1 K

To find Z12 and Z22 let I1 be zero by making port 1 1 | open circuited.

eq Z V

I 22 = 310)12(10

X = 3.33 m A.

V1 = I2 X Z3 = 3.33 X 10 3 X 1 X 10 3 V1 = 3.33 V

Z12 = 2

1

I V

at I1 = 0 = 31033.333.3

X = 1 K

Z22 = 2

2

I

V at I1 = 0 =

31033.3

10

X = 3 K

Z11 = 4 K , Z12 = 1 K , Z21 = 1 K , Z22 = 3 K


41/73



Y PARAMETERS: I1 = Y11 V1 + Y12 V2; I2 = Y21 V1 + Y22 V2

To find the values of Y11 and Y21, let V2 be zero by making short circuit at port 2 2 |

Req = 3 + 2 | | 1 = 3.667 K .

I1 = eq R

V 1 = 310667.3

10 X

= 2.73 m A

I2 = I1 X 32

3

Z Z

Z = 2.73 X 10 3 X 3

3

103101

X X

= 0.91 m A

Y11 = 1

1

V I

at V2 = 0 = 10

1073.2 3 X = 0.273 m

Y21 = 1

2

V I

at V2 = 0 = 10

1091.0 3 X = 0.091 m

To calculate Y12 and Y22 make V1 at 1 1 | port is zero. Zeq = 2 + 3 | | 1 = 2.75 K

eq Z V

I 22 = 31075.210 X

= 3.636 m A

I1 = I2 X 31

3

Z Z

Z = 3.64X10 3 X 3

3

104101

X X

= 0.91 m A

Y12

=

2

1

V

I

at V1

=

0

=

10

10091.0 3 X =

0.091

m

Y22 = 2

2

V I

at V1 = 0 = 10

10636.3 3 X = 0.364 m

Y11 = 0.273 m , Y12 = 0.091 m , Y21 = 0.091 m , and Y22 = 0.364 m

TABULAR FORMS: Z PARAMETERS:

V1 = 10 V, I2 = 0 A (O.C) V2 = 10 V, I1 = 0 A (O.C)

Sl. No.

Parameter Theoretical Practical Sl.

No. Parameter Theoretical Practical

1 V2 2.5 V 1 V1 3.33 V

2 I1 2.5 mA 2 I2 3.33 mA

3 4 k 3 1 k 4 1 k 4 3 k


42/73



Y PARAMETERS:

V1 = 10 V, V2 = 0 A (S.C) V2 = 10 V, V1 = 0 A (S.C)

Sl.


Sl.


1 I1 2.73 mA 1 I2 3.63 mA

2 I2 0.91 mA 2 I1 0.91 mA

3 0.273 m 3 0.091 m 4 0.091 m 4 0.364 m

PROCEDURE:

1. Connections are made as per the circuit diagram. 2. To find the values of Z11, Z21, open circuit at port 22I to make I2 = 0, and Note down

the values of I1 and V2 3. To find the values of Z12 and Z22, make the current I1 = 0 by open circuiting the

terminals of port 1 1I and the note down the values of I2 and V 1 4. To find the Y parameters, making short circuit at port 2 2I i.3. V2 = 0 and note

down the values of I1 and I2 to determine Y11 and Y21 parameters 5. Short circuit at port 1 1I i.e. V1 = 0 to find the values of Y12 and Y 22 with the

readings of I1 and I2. 6. From the obtained readings calculate all Z and Y parameters and verify with

Theoretical values.

RESULT:

Open circuit impedance (Z) parameters and Short circuit (Y) admittance parameters for the two port network are verified.


43/73




1. Explain two port networks.2. What are Z & Y parameters?3. Why Z parameters for series impedance network does not exist?4. Why Y parameters for parallel impedance network does not exist?5. What are the applications of Two port networks?6. Define port.7. Define passive ports.8. Define active ports.

9.

What is the driving point impedance at port 1 with port 2 opencircuited for the given network.

10. What is the advantage of two port analysis?

21

2

11

1

321


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Expt. No: 07Date:

SERIES AND PARALLEL RESONANCEAIM:

To find the resonant frequency, band width and quality factor inseries and parallel circuits.

APPARATUS:

Sl. No. DESCRIPTION SPECIFICATION QUANTITY

1 Function Generator 1MHZ 1

2 Cathode ray Oscillator 20MHZ 1

3 Decade Resistance Box 1

4 Decade Inductance Box 1

5 Decade Capacitance Box 1


7 Bread board WB-102 1

8 Connecting wires L.S.

CIRCUIT DIAGRAM:

SERIES RESONANCE:

PARALLEL RESONANCE:


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THEORY:SERIES RESONANCE:

RESONANT FREQUENCY:In any electrical circuits, resonance is very important phenomenon.

In a series RLC circuits, the current lags behind or leads the applied voltagedepending upon the values of X L and X C. When X L > X C, the circuit ispredominantly inductive and the current lags behind the voltage and whenXL < X C, the current is predominantly capacitive and the current leads thevoltage. However if one of the parameters in RLC circuit is varied in such away that the current in the circuit is in phase with the applied voltage, thenthe circuit is said to be (at resonance) resonating.

Total impedance in RLC series circuit isZ = R + j(X L - X C) = R + j ( L 1/ C)

It is clear from the circuit that the current I = V/Z

In a series RLC circuit, series resonance occurs when X L =X C. Thefrequency at which the resonance occurs is called the resonant frequency.Since X L = X C the impedance in series RLC circuit is purely resistive at theresonant frequency f r (XL = X C. i.e. L = 1/ C)

2 f rL =C f r 2

1 r f 2 = LC 24

1 r f =

LC 2

1

Where f r = Resonant frequencyAt resonant frequency, the capacitive reactance is equal to inductive

reactance, and hence the impedance is minimum. Because of minimumimpedance, maximum current flows through the circuit, the currentvariation with frequency is plotted.BANDWIDTH:

The bandwidth of any system is the rangeof frequencies for which the current onvoltage is equal to 70.7% of its value atthe resonant frequency and it is denotedby band width.

Here the frequency f 1 is thefrequency at which the current is 0.707times the current at resonant value and itis called the Lower cut-off frequency.

The frequency f 2is the frequency atwhich the current is 0.707 times thecurrent at resonant value and is calledthe Upper cut-off frequency.

The Bandwidth or B.W is definedas the frequency difference between f 2and f 1

Lower cut-off frequency Upper cut-off frequency Bandwidth =

QUALITY FACTOR:It is known as figure of merit and is an indication of the quality of the

coil.

or or A higher value of Q results in smaller bandwidth and a lower value ofQ causes higher bandwidth.


46/73



CONDITIONS AT RESONANCE IN SERIES CIRCUIT:1. The inductive and capacitive reactances are equal.2. The net reactance is zero.3. Impedance is minimum and is equal to Resistance.4. The current is maximum and is equal to V/R.5. The V and I are in phase and power factor is unity.

PARALLEL RESONANCE: The parallel circuit consisting of energy storage elements (L and C)

behaves as a resistive network at resonance and the applied voltage and theresulting current is in phase.

In parallel circuit, the net Susceptance is zero at resonance.sin sin = = Finally, Resonant frequency When R L = RC, Resonant frequency r f =

LC 2

1

CONDITIONS AT RESONANCE IN PARALLEL CIRCUIT:1. The inductive and capacitive susceptances are equal.2. The net susceptance is zero.3. Admittance is minimum and is equal to Conductance.4. The current is minimum at resonance.5. The V and I are in phase and power factor is unity.

THEORITICAL CALCULATIONS:SERIES RESONANCE:

r f = LC 2

1 =

63 101.0102002

1

X X X = 1,125.39 Hz. = 1.125 k Hz.

Lower cut-off frequency 1 4 = 1,125.39 = 1,125.39 19.89 = 1105.5 Hz.Upper cut-off frequency 2 4 =

1,125.39 = 1,125.39 + 19.89 = 1145.28 Hz.

Bandwidth = 2 1 = 1145.28 - 1105.5 = 39.78 Hz.Quality factor =12 f f

f r = .. . = 28.29PARALLEL RESONANCE:

In a parallel RLC circuit the resonant frequency, ,When R L = R C, the Resonant frequency is given by r f =

LC 2

1

Resonant frequency, . = 2250.79 Hz., 2.25 k Hz.


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MODEL GRAPH:

PROCEDURE:

SERIES RESONANCE: Connections are made as per the circuit diagram shown in fig. By varying the Input Frequency, note down the corresponding value

of current. Plot the graph between Frequency and current for the series circuit. Calculate the band width and quality factor to the circuit.

PARALLEL RESONANCE: Connections are made as per the circuit diagram shown in fig. By varying the Input Frequency, note down the corresponding value

of current. Plot the graph between Frequency and current for the parallel circuit. Calculate the Resonant frequency for the circuit.

OBSERVATIONS:SERIES RESONANCE PARALLEL RESONANCE

Sl.No.

Frequency(kHz)

Current(mA)

Sl.No.

Frequency(kHz)

Current(mA)

RESULT:

Resonant Frequency, Bandwidth and Quality factor is calculated forboth Series and Parallel resonating Circuits.


48/73



VIVA VOCE QUESTIONS:

1. What is meant by Resonance?2. What is the condition for resonance in series RLC circuit?3. Why the RLC series circuit is called Acceptor circuit and RLC

parallel circuit is called Rejecter circuit?4. Define Q-factor and Bandwidth.5. Expressions for frequency of Resonance in series and parallel RLC

circuits. 6. What is the total reactance of a series RLC circuit at resonance? 7. What is the phase angle of a series RLC circuit at resonance?8. Define resonant frequency. 9. Give the expression for frequency at resonance in series RLC

circuit. 10. What is the power factor of series and parallel RLC resonant

circuit?


49/73



Expt. No.: 08Date:

TRANSMISION & HYBRID PARAMETERSAIM:

To find the Transmission and Hybrid parameters of a given network.APPARATUS REQUIRED:Sl.No.


1 Transistorized Power Supply (0-30)V 1

2 Resistance 1k 1

3 Resistance 2k 1

4 Resistance 3k 1

5 Ammeter (0 - 200) mA 1

6 Bread board WB - 102 1



CIRCUIT DIAGRAMS

TRANSMISSION PARAMETERS:

I2=0 i.e., output port open circuited to find A, C. V2=0 i.e., output port short circuited to find B,D.

HYBRID PARAMETERS:

V2=0, output port short circuited to find h11 , h21 . I1=0, input port open circuited to find h12 , h22.


50/73



THEORY: A two port network is simply a network inside a black box, and the network has only

two pairs of accessible terminals. A port is defined as any pair of terminals into which

energy is supplied, or from which energy is withdrawn. Usually one pair represents the

input and the other represents the output. Such a building block is very common in electronic systems, transmission and distribution systems.

Figure shows a two port network, or two terminal pair network, in which the four terminals have been paired into ports 1 1 | and 2 2 | . The terminals together 1 1 | together constitute a port. Similarly the terminals 2 2 |

constitute another port.

The voltage and current assigned to each of the two ports. The voltage and current

at the input terminals are V1 and I1 where as V2 and I2 are specified at the output port. It is

also assumed that the currents I1 and I2 are entering into the network at the upper terminals 1 and 2 respectively. The variables of the two port network are V1, V2 and I1, I2.

Two of these are dependent variables the other two are independent variables.

TRANSMISSION PARAMETERS: The Transmission parameters of a two port

for the positive directions of voltages and currents may be defined by expressing the input port variables V1 and I1 in terms of the output variables V2 and I2. Here V1 and I1 are dependent variables and V2 and I2 are independent variables.

(1) (2) Negative sign indicates the Reverse current direction.

Thus

The matrix is called the Transmission matrix

The individual A, C parameters for a given network can be defined by setting output port current is equal to zero, suppose port 2 2 I is left open circuited then I2 = 0

0 is called Open circuit voltage gain 0 is the open circuit Transfer impedance.

Output port 2 2 I is short circuited to make V2 = 0

0 is the Short circuit Transfer Admittance

0 Short circuit Current gain


51/73



HYBRID PARAMETERS: The Hybrid parameters of a two port for the

positive directions of voltages and currents may be defined by expressing the voltage at the input port V1 and the current at the output port I2 in terms of

I1 and V2. Here V1 and I2 are dependent variables and I1 and V2 are independent variables. (1) (2)

Thus Output port 22I is short circuited to make V2 = 0 0 Short circuit input impedance

0 Short circuit forward current gain

Input port 11I is left open circuited to make I1 = 0

0 Open circuit Reverse voltage gain 0 Open circuit output Admittance

THEORITICAL CALCULATIONS: TRANSMISSION PARAMETERS: When

I2

= 0,

i.e.

2

2

I is

open

circuited.

eq RV

I 11 = 310)13(10

X = 2.5 m A

V2 = I X 1 X 10 3 = 2.5 X 10 3 X 1 X 10 3 = 2.5 V

0 102.5 4 0 2.5 1010 0.25 To find B and D let V2 be zero by making output port 22I is short circuited. Req = 3 + 2 | | 1 = 3.667 K .

I1 = eq R

V 1 = 310667.3

10 X

= 2.73 m A

I2 = I1 X 32

3

Z Z

Z = 2.73 X 10 3 X 3

3

103101

X X

I2 = 0.91 m A

0 100.91 10

10.99 0 2.73 100.91 10 3


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A = 4; B = 10.99 k ; C = 0.25 m ; D = 3 HYBRID PARAMETERS:

To find the values of h 11 and h21 , let V2

be zero by making short circuit at port 2 2 |

Req = 3 + 2 | | 1 = 3.667 K .

I1 = eq R

V 1 = 310667.3

10 X

= 2.73 m A

I2 = I1 X 32

3

Z Z

Z = 2.73 X 10 3 X 3

3

103101

X X

= 0.91 m A

0 102.73 10 3.66 0 0.91 10

2.73 10 0.33

To calculate h12 and h22 make I1 at 1 1 | port is zero.

eq Z V

I 22 = 310)12(10

X = 3.33 m A.

V1 = I2 X Z3 = 3.33 X 10 3 X 1 X 10 3 V1 = 3.33 V

0 3.3310 0.333

0 3.33 1010 0.333 h11 = 3.66 k h12 = 0.333, h21 = 0.33 and h22 = 0.333 m

TABULAR FORMS: TRANSMISSION PARAMETERS:

V1 = 10 V, I2 = 0 A (O.C) V1 = 10 V, V2 = 0 V (S.C)

Sl.

No. Parameter

Theoretical

Practical

Sl.

No. Parameter Theoretical

Practical

1 V2 2.5 V 1 I1 2.73 mA

2 I1 2.5 mA 2 I2 0.91 mA

3 4 3 10.99 k

4 0.25 m 4 3


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HYBRID PARAMETERS:

V1 = 10 V, V2 = 0 A (S.C) V2 = 10 V, I1 = 0 A (O.C)

Sl.


Sl.


1 I1 2.73 mA 1 I2 3.33 mA

2 I2 0.91 mA 2 V1 0.33

3 3.66 k 3 0.333 4 0.33 4 0.33 m

PROCEDURE:

1. Connections are made as per the circuit diagram.

2. To find the values of A, C open circuit at port 22I to make I2 = 0, and Note down the

values of I1 and V2

3. To find the values of B and D, make the voltage V2 = 0 by short circuiting the

terminals of port 2 2I and the note down the values of I1 and I 2

4. To find the h parameters, making short circuit at port 2 2I i.3. V2 = 0 and note

down the

values

of

I1

and I2

to

determine

Y11

and Y21

parameters

5. Open circuit at port 1 1I i.e. I1 = 0 to find the values of h12 and h22 with the readings

of V1 and I2.

6. From the obtained readings calculate all Transmission and h parameters and verify

with Theoretical values.

RESULT:

Transmission parameters and hybrid parameters for the two port network are

verified.


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1. Explain two port networks.2. What are Transmission & Hybrid parameters?3. What are the applications of Two port networks?4. Define port.5. Define passive ports.6. Define active ports.7. What is the driving point impedance at port 1 with port 2 open

circuited for the given network.

8. What is the advantage of two port analysis?

21

2

11

1

321


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Expt. No.: 09

Date:

TIME RESPONSE OF FIRST ORDER RC NETWORK AIM:

To find the time response of first order RC network for Non-Sinusoidal input (Square Input).

APPARATUS REQUIRED:

CIRCUIT DIAGRAM:

HIGH PASS RC CIRCUIT LOW PASS RC CIRCUIT

THEORY: Time response means the response (or) variation of output voltage

with change in time. Here, we consider the RL and RC circuits.Low pass circuits are those which rejects all frequencies above aspecified value called cut off frequency. The signal of all frequencies ispassed without attenuation.

The High pass circuits are those which rejects all frequencies below aspecified value. This circuit is a compliment of Low pass circuit. This passesall frequencies above the cut off frequency without attenuation.TIME CONSTANT:

Time constant is defined as the time during which voltage across thecapacitor would have reached to its maximum value v and it is maintainedits initial rate of rise = RC, the time constant for RL circuit is given by =L/R.

Sl. No. Description Range &Type Quantity

1 Function generator 1

2 Decade resistance box 1 ohm 10 M ohm 1

3 Decade capacitance box 100 pf 1 micro f 1 4 Cathode Ray Oscilloscope 1

5 Bread Board 1

6 Patch cards L.S.


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These RL and RC circuits are useful in many applications. RCdifferentiator finds useful applications in converting a triangular intosquare wave. By using High pass RC circuits pulses can be considered intospikes by making time constant small.

Low pass circuit can be used as a voltage sweep generator (or) rampgenerator. We obtain under damped or over damped responses using RLCcircuits. We can obtain average or DC values of output voltages at constant,whatever the DC value of input voltage using the circuit.

We can study a band of frequencies using these circuits. In electricalengineering Low pass and High pass filters are also utilized in order toeliminate the undesired frequency components resulting from controlcircuits.RL CIRCUIT (STEP INPUT):

V dt di

L Ri (or) LV

i L R

dt di

L R

D i = LV

Since it is a Non Homogeneous linear equation it (current, i) has two parts(i) = i c +ip

ic = Ct

L

R

e Amp.

ip =

L R

D

e LV t 0

put D = 0, i p = R L

X LV

= RV

ip = RV

pc iii = Ct

L

R

e

+ RV

At t = 0 i (0) = 0 RV

C

Then i = RV

t L

R

e + RV

=

t

L

R

e RV

1 =

t

e RV

1 amps Here = R L

voltage across resistor V R = iR

VR = R X

t

e R

V 1

= V

t

e1V


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Voltage across inductor V L =dt di

L

VL = L R

X RV

L

t

e = V

t

e V

RC CIRCUIT (STEP INPUT)

idt C

RiV 1

By differentiating the above equation, we get

iC dt

di R

1 = 0 (or)

dt di

+ R

1 i = 0

Since the equation is a Homogeneous linear differential equation. Ithas only complimentary solution.

t RC Cei

1

ampAt t = 0

RV

i substitute in the above equation, RV

C

RV

i t

RC e

1

(or)

t

e RV

i Here = RC sec.Voltage across resistor R is

Rv R X i = R X

t

e RV

Rv = V

t

e --------- 3

Voltage across capacitor C is

C v = idt C 1

= RV

C 1

t

RC Ce

1

dt = RC V

X -1

RC RC

t

e + C = - V t

e + C

At t = 0 voltage across capacitor is VC v = 0 = - V e 0 + C C = V

C v = V

t

e1 V -------- 4


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HIGH PASS RC CIRCUIT

Consider the RC circuit shown in figure.

Let V in = Input signal andV0 = Output voltageI = Current flows through the circuit.

If the input signal Vi is non-sinusoidal, it can be visualized ascomprising of several sine waves of frequencies which are multiples of thefrequency of the signal, and there may be a d.c. component also. Since thereactance of the capacitor decrease as the frequency increases Xc=1/2 fC,the capacitor offers very little impedance to harmonics of higher orders.Hence the high frequency components of the input voltage pass through thenetwork with very little attenuation. For this reason the circuit is termed asHigh Pass RC circuit.

LOW PASS R-C CIRCUIT

Consider the RC circuit shown in figure

Let V in = Input signal andV0 = Output voltageI = Current flows through the circuit.

It may be observed that this circuit is same as the high pass RC

circuit, so far as the circuit configuration is concerned. However there isone basic difference, where as in the high pass circuit the output is takenacross the resistor, in the circuit under consideration the output is takenacross the capacitor.

Since the reactance of a capacitor decreases with increase infrequency and vice versa it can be seen that the capacitor offers largerimpedance to the low frequency components of the input voltage and hencethese low frequency components pass out easily to the output without anyappreciable attenuation. The circuit is therefore called Low Pass RCcircuit

RESPONSE OF HIGH PASS RC CIRCUIT TO SYMMETRICAL SQUARE WAVE INPUT

Consider the high pass RC circuit as shown in fig. Let the forcing

functio9n be a symmetrical square wave of peak to peak amplitude V andperiod T


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Case (i) Let RCT

If the time constant is quite large the

exponential decay of the output voltage isslower, as seen already. This results in atilt at the top of the input wave and anundershoot at the bottom, as is obviousfrom the output wave form.

PQ is the tilt at the top of the wave. Thegeneral expression for percentage tilt is

% Tilt = RC

T

RC

T

e

e

2

2

1

1 X 200

If RC >> T The percentage tilt becomes. % Tilt = RC T

2 x 100

RESPONSE OF LOW PASS RC CIRCUIT TO SYMMETRICALSQUARE WAVE INPUT

Case (i) Let = RC


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Case (ii) Let = RC >> TAt t = 0 the input voltage is at V/2 and theoutput voltage is exponentially increasing and itcannot reaches to its maximum value at t = T/2because of large time constant. When t = T/2the input voltage is (V/2), then the outputexponentially decreases and the voltage waveform is repeats for the successive input wave.

The output wave forms when RC >> T is shownin figure.


Case (i) At RC = = tR = 1 K , C = 1 F, T = 1 ms. = 1 x 10 3 x 1 x 10 -6 s.

V0 =2V tanh x where x =

RC T

4

AT V = 20 V, R = 1x10 3 , C = 1x10 -6 F,t = 1x10 -3 s

V0 =2

20tanh 3

3

1014101 X X

X = 2.44 V

Case (ii) At RC = > tFor = 10 m sec, t = 1 m sec,R = 1 k (between 1 k 10 k ), C = 1 F.

V0 =2V

tanh x where x = RC T

4

V0 = 220 tanh 33

10104101 X X

X = 0.25 V

PROCEDURE:

1. Connect the circuit diagram as per the circuit to observe the outputvoltage across resistor.

2. Apply Non sinusoidal square input of 20 V peak to peak and with atime period (t) of 1 m sec.

3. Observe the wave form across R at different conditions viz, (1) = t,(2) > t, by varying its Time constant where = RC

4. Draw the observed wave forms on a graph sheet


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Expt. No.: 10Date:

MEASUREMENT OF ACTIVE POWER FOR STAR CONNECTED

BALANCED LOADAIM: To measure the Active power for Star connected balanced load.

APPARATUS:Sl.No.

Apparatus Type & range Quantity

1 Volt meter (0 - 600) V MI 1

2 Ammeter (0 - 5) A MI 1

3 Watt meter 600 V,2.5 A LPF 2 4 Inductive load 415 V/10A 1

5 3- Auto

transformer (0 - 470) V/10A 1

CIRCUIT DIAGRAM :

THEORY:

The relationship between real power, reactive power and apparentpower can be expressed by representing the quantities as vectors. Realpower is represented as a horizontal vector and reactive power isrepresented as a vertical vector. The apparent power vector is the

hypotenuse of a right triangle formed by connecting the real and reactivepower vectors. This representation is often called the power triangle . Using


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the Pythagorean Theorem , the relationship among real, reactive & apparentpower is:

Real and reactive powers can also be calculated directly from theapparent power, when the current and voltage are both sinusoids with aknown phase angle between them:

The ratio of real power to apparent power is called power factor and isa number always between 0 and 1. Where the currents and voltages havenon-sinusoidal forms, power factor is generalized to include the effects ofdistortion

TWO WATT METER METHOD:

If the load is balanced, then power factor of the load can also befound from the two wattmeter readings. Let V R, V Y, V B be the r.m.s. valuesof the three phase voltage and I R, I Y, I B be the r.m.s. values of the threephase currents. Since the base voltages and currents are sinusoidal, theycan be represented by vectors, the current lagging behind their respectivephase voltages by .

Current through wattmeter 1 W 1 is = I R.Potential difference across voltage coil of W 1 is V RB = V R V B. This V RB founded by compounding V R and V B reversed. It is seen that

phase difference between V RB and I R = (30 ).

cos30 Similarly the current through wattmeter 2 W 2 is = I Y Potential difference across voltage coil of W 2 is V YB = V Y V B.Again V YB is found by compounding V Y and V B reversed. The angle

between I Y and V YB is = (30 + ).

cos30 .

Since load is balanced, V RB = V YB = Line voltage V L , and I R = I Y = Linecurrent, I L .

cos 30 cos 30 cos 30 cos 30 cos30cos sin30si 2 cos30cos 2 cos 3 cos = Total Power in the 3 phase load.


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VRY = 415 0 V,VYB = 415 120 V,VBR = 415 240 V, 3 30 415 3 0 30 239.6 30 R = 330 , X L = j 1040 ,

ZR = (330+j1040) = 1091.1 72.39 239.6 01091.172.39 0.2196 72.39 cos 3301091.1 0.302 Total Active Power (P) = 3 x V Ph x I Ph x cos = 3 x V R x I R x cos

(P) = 3 x 239.6 x 0.2196 x 0.302 = 47.67 W (OR) Total Active Power (P) = 3 I 2 R = 3 x 0.2196 2 x 330 = 47.52 W

PROCEDURE:

1. Connect the circuit as per circuit diagram. 2. Apply the suitable voltage to the circuit by using Auto Transformer 3. Note down the reading of the Watt meters, Ammeter and Voltmeter. 4. Calculate the Active power consumed by the load using given

formulae.

TABULAR FORM:

Sl.No.

Voltage(Volt)

Current(Amp)

W1 inWatt

W2 inWatt

Active Power (P)

TheoreticalP in Watt

P = W 1+W 2in Watt

1 415 0.37 68 - 20 47.67 48

RESULT: The amount of Active Power in a 3 phase balanced system ismeasured by using Two watt meter method


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VIVA VOCE QUESTIONS:

1. What is Apparent Power2. Formula for the Active Power

3. Unit of Active power

4. What is difference between active and reactive power

5. Explain the power triangle

6. What meant by M,L,C,V in wattmeter

7. Explain the types of wattmeter

8. What is the difference between LPF and UPF wattmeter9. When the LPF wattmeter is used in the circuit

10. Which is the best wattmeter to use in both AC and DC sources


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Expt. No.: 11Date:

22 Ec Lab Master Manual 14

Documents

Transcript of 22 Ec Lab Master Manual 14