Multisim/Networks Lab

Department of Electrical and Electronics Engineering

MULTISIM / NETWORKS

Laboratory Manual

GOKARAJU RANGARAJU INSTITUTE OF ENGINEERING AND TECHNOLOGY

(Autonomous Institute under JNTU Hyderabad)

MULTISIM / NETWORKS LAB

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CERTIFICATE

This is to certify that it is a bonafide record of practical work done in the Multisim/Networks Laboratory in I sem of II year during the year

2011-2012 Name: Roll No: Branch: Signature of staff member


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Contents

1.Thevenin’s Theorem.

2. Norton’s Theorem

3. Maximum Power Transfer Theorem.

4. Superposition and Reciprocity Theorems.

5. Z and Y parameters.

6. Transmission and Hybrid Parameters.

7. Compensation and Milliman’s Theorems.

8. Series Resonance

9.Parallel Resonance.

10. Locus of Current Vector in an R-L Circuit

11. Locus of Current Vector in an R-C Circuit

12Measurement of 3-phase power by two wattmeter method for unbalanced loads.

13. Measurement of Active and Reactive power by star and delta connected balanced loads.


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1. Thevenin’s Theorem

Aim:

1. To construct a circuit and verify Thevenin’s Theorem for the given circuit.

Apparatus Required:

1. Voltmeter

2. Resistances

3. Bread board

4. Ammeter

5. DC voltage source

Theory:

Thevenin’s Theorem:

This theorem states that a network composed of lumped, linear circuit elements may , for the

purposes of analysis of external circuit or terminal behaviour, be replaced by a voltage source

V(s) in series with a single impedance.

Thevenin’s theorem simplifies the method of finding current through any specified branch. For this

purpose we have to find two things:

1. Thevenin’s Resistance Rth

2. Thevenin’s Voltage Vth


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Circuit Diagram:

Theoretical Calculations:

To find current through 1k ohm resistor using Thevenin’s theorem:

1) To find Thevenin’s resistance (Rth) across 1k ohm resistor:

R1

2.2k

R2

2.2k

0

1

Rth = (2.2* 2.2)*106/ (2.2+2.2)(10)

3= 1.1k ohm

2) To find Thevenin’s voltage (Vth) across 1k ohm resistor:

R1

2.2k

R2

2.2k

1

V110 V

2

00

I=10/4.4*10

3 =2.27mA


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Applying KVL,

-10 + (2.2*103*2.27*10

-3) +Vth =0

Vth=5.006V Thevenin’s equivalent circuit is:

R1

1.1k

V15.006 V

2

0

1

Finding current through 1k ohm resistor using Thevenin’s theorem,

R1

1.1k

R2

1k

V15.006 V

2

0 0

3

It=5.006/ (2.1*103) = 2.38 mA

Current through 1k ohm resistor is 2.38mA.

Hence Thevenin’s theorem is verified.

Procedure:

A) Thevenin’s procedure

1. Remove the resistor R5.4. Remove the voltage source and short the terminals 2, 4.

5. Resistance measured between 1, 3 is Thevenin’s resistance.

6. Thevenin equivalent circuit is obtained by connecting Vth and Rth in series.

7. Connect the resistance 1K in series with Thevenin equivalent circuit and measure current across the load

8. Verify the current measured in Thevenin equivalent circuit and original circuit.

Observations:

Thevenin’s Voltage (Vth) =


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Thevenin’s Resistance (Rth) =

Load Current (IL) =

Multisim Results:

Thevenin’s Voltage (Vth) =

Thevenin’s Resistance (Rth) =

Load Current (IL) =

Theoretical Calculations to be done by Students:


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Result:

1. Thevenin’s theorem is verified.


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2. Norton’s Theorem

Aim:

To construct a circuit and verify Norton’s Theorem for the given circuit.

Apparatus Required:

1. Voltmeter

2. Resistances

3. Bread board

4. Ammeter


Theory:

Norton’s Theorem:

Any linear circuit containing several energy sources and resistance can be replaced by a single constant generator in

parallel with a single resistor.

Circuit diagram:

R1

100

R2

150

R351

V110 V

1

3

2


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Norton’s procedure:

1. Remove the resistance R2.

2. Insert an ammeter across the open terminals.

3. Measure the resistance between the terminals replacing 10v DC source with a ‘short’ let us say this equals

Rn (Norton’s resistance)

4. Construct an equivalent circuit and verify the current across the load in both circuits.

Theoretical calculations:

R1

100

R2

150

R351

V110 V

1

3

2

STEP 1:

Finding R equivalent:

To find R

R1

100

R251

0

1


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Req= (100*51)/151

=33.77ohm

STEP 2:

To find IN:

R1

100

R251

V110 V

1 4

Since there is a short circuit path across R2, so current will not pass through R2, R2 can be neglected.

IN=10/100=0.1A

I10.1 A

R133.77 R2

150

1

2

I150= (0.1)*(33.77/33.77+150)

=0.0183A

Observations:

Norton’s Current (IN) =


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Norton’s Resistance (RN) =

Load Current (IL) =

Multisim Results:

Norton’s Current (IN) =

Norton’s Resistance (RN) =

Load Current (IL) =



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Result:

Norton’s theorem is verified.


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3. Maximum Power Transfer Theorem

Aim:

1. To construct a circuit and verify Maximum Power transfer Theorem for the given circuit.

Apparatus Required:

1. Voltmeter

2. Resistances

3. Bread board

4. Ammeter


Theory:

Maximum Power transfer theorem:

Maximum power transfer theorem states that the power delivered from a source to a load is maximum when source

resistance equals load resistance.

Circuit Diagram:

V110 V

R1

2.2k

RL

R32.2k

1 a

0

Procedure:

Maximum Power transfer theorem


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1. Construct the circuit.

2. Connect the circuit with different loads.

3. Note down the power delivered to load and voltage.

4. Verify the resistance at which maximum power is delivered is equal to R1.

Observations:

For Maximum power transfer:

S.No V(volts) I(mA) Power delivered

To load V×I

R(load)Ώ V2/4RL

1. 2.381 2.381 5.66m 1k

2. 2.5 2.273 5.68m 1.1k 5.68

3. 3.33 1.515 5.04m 2.2k

4. 1.687 3.012 5.08m 2.2k

5. 0.417 4.167 1.73m 100

6. 4.051 0.863 3.49m 4.7k

Maximum power transfer calculations:

Load current= I= VS / (RN + RL)


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P= Power delivered to load = (VS / (RN + RL)) 2 RL

Maximum power transferred = V2/4RL

STEP1:

To find equivalent resistance across ab (Rab)

1.Rab is found out by shorting the voltage source and calculating resistance across a&b.

Rab= ((2.2K)//(2.2k))/94.4k)

=1.1k Ώ

STEP 2:

Finding VTH

1. VTH is the voltage across a&b.

V210 V

R5

2.2k

R62.2k

2

4

0

R2

2.2k

R42.2k

3

0


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Voltage across ab is VTH

Current through 2.2k=10/ (2.2+2.2) k

=2.27mA

VTH= (2.2k)*(2.27m)

=5V

Therefore, VTH=5V

Maximum power transfer occurs when RL=Rab=1.1KΏ

Power transferred = (VTH*VTH)/ (4*RL)

=25/ (4*1.1k)

=0.00568W.

Also, V*I= (2.5)*(2.2727m)

=0.00568W

Therefore VI= (VTH*VTH)/ (4*RL)

Hence maximum power transfer theorem is verified.

Theoretical calculations to be done by students:


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Result:

Maximum power transfer theorem is verified.


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4. Super position Theorem and Reciprocity Theorem.

Aim:

1. To construct a circuit and verify Super position Theorem for the given circuit.

2. To construct a circuit and verify Reciprocity Theorem for the given circuit.

Theory: A) Super position Theorem:

In any linear network containing two or more sources ‘response’ in any element is equal to the algebraic

sum of responses caused by individual sources acting while the other sources are inoperative.

The word inoperative means a voltage source is replaced by a short circuit while the current source

replaced by open circuit.

B) Reciprocity Theorem:

In a circuit having several branches, if a source of voltage V produces a current I in another branch, the

same current I will flow in the first branch if voltage source is put in the second branch. That means

voltage source and ammeter can be interchanged but the ammeter reading will remain unaltered.

Circuit Diagram:

A) Super position Theorem:

.

B) Reciprocity Theorem:

R1

2.2k

R2

4.7k

R3 3.3k

V1 10 V

1

V2 8 V

10

0

2


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R13.3k

R2

2.2k

R31.0k

R4

10k

R54.7k

V110 V

2 3

4

0

U1DC 1e-0090.099m A

+

-

5

6

Procedure:

Superposition:

1. First measure the current through R5 due to source V1 while source V2 is replaced with short circuit. Let

this current be Iv1.

2. Next measure current through R5 due to source V2 while source V1 is replaced with short circuit.

3. Let this current be Iv2.

4. Now let both sources be in place. The current through R5 is measured once again. Let this current be I.

5. Verify whether I= Iv1+Iv2.

Reciprocity:

1. Construct the circuit given.

2. Measure the current in the R5.

3. Now replace ammeter with voltage source and voltage source with ammeter measure the current in R3.

4. Compare both readings.


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Observations:

1) Superposition theorem:

V1

(volts)

V2

(volts)

R1

(ohms)

R2

(ohms)

R3

(ohms)

I1

(mA)

I2

(mA)

I’3.3k

=

I’’3.3k

(mA)

10 2 2.2k 4.7k 3.3k 1.4194 0.5316 1.951

2) Reciprocity Theorem Experiment:

R1

(kohms)

R2

(kohms)

R3

(kohms)

R4

(kohms)

R5

(kohms)

Vs

(volts)

I

(mA)

I’

(mA)

Vs/I

(kohms)

Vs/I’

(kohms)

3.3 2.2 1 10 4.7 10 0.09 0.09 111.11 111.111




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Consider 10V D.C. voltage source and replace 8V D.C. voltage source with short circuit.

R1

2.2k

R2

4.7k

R33.3k

V110 V

1

U2DC 1e-0091.419m A

+

-

4

2

0

Total resistance, RT = (4.7k||3.3k) +2.2k

=4.139 kohms.

Total current, I = 10 / RT

=10 / 4.139

=2.416 mA.

The current through 3.3kohm resistor is,

I’= I x 4.7k / (4.7k + 3.3k)

= 2.416 x 4.7 / 8k

R1

2.2k

R2

4.7k

R3 3.3k

V1 10 V

1

V2 8 V

10

0

2


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= 1.4194 mA.

Now, consider 8V D.C. voltage source and replace 10V D.C. voltage source by short circuit.

R42.2k

R53.3k

R6

4.7k

V28 V

5

U1DC 1e-0090.531m A

+

-

7

0

8

Total resistance, RT = (2.2k || 3.3k) + 4.7k

= 6.02 kohms.

Total current, I = 8 / 6.02k

= 1.3289 mA.

The current through 3.3kohm is,

I2 = 1.3289mA x 2.2k / 5.5k

= 0.5316 mA.

Therefore, the total current passing through 3.3kohm

I’3.3k = I1 + I2

=1.4194 + 0.5316

= 1.951 mA.

Now consider both the voltage sources,

V’


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V310 V V4

8 V

R7

2.2k

R83.3k

R9

4.7kU3DC 1e-0091.950m A

+

-

3 9

11

6

0

Applying nodal analysis,

(V’ – 10)/2.2k + V’/3.3k + (V’-8)/4.7k =0

Therefore, V’ = 6.4388V.

The current through 3.3kohm resistor is,

I’’3.3k = V’ / 3.3k

= 6.4388 / 3.3k

= 1.9512mA.

HENCE PROVED.

2) Reciprocity theorem:

V’


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V’ /5.5k + V’/1k + (V’ – 10)/14.7k =0

V’ = 0.54V.

The current I in the 3.3kohm resistor branch is,

I = V’ / 5.5k = 0.54/5.5k =0.09mA.

Now, the reciprocal circuit to the above circuit is,

V’

R1 3.3k

R2

2.2k

R3 1.0k

R4

10k

R5 4.7k

V1 10 V

2 3

4

0

U1 DC 1e-009 0.099m A

+

-

5

6


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U1DC 1e-0090.099m A

+

-

V110 V

R13.3k

R2

2.2k

R31.0k

R4

10k

R54.7k

1

2 3

46

0


(V’ – 10) / 5.5k + V’ / 1k + V’ / 14.7k =0

V’ = 1.45V.

The current I in the branch is,

I’ = V’ / R = 1.45 / 14.7k = 0.09mA.

HENCE PROVED.

Bread Board Results:


V1

(volts)

V2

(volts)

R1

(kohms)

R2

(kohms)

R3

(kohms)

I1

(mA)

I2

(mA)

I’3.3k

=

I”3.3k

(mA)

10 8 2.2k 4.7k 3.3k 1.42 0.5 1.92


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Vs

(volts)

I

(mA)

I’

(mA)

Vs / I

(kohms)

Vs / I’

(kohms)

10 0.1 0.1 100 100

Multisim Results:


V1

(volts)

V2

(volts)

R1

(kohms)

R2

(kohms)

R3

(kohms)

I1

(mA)

I2

(mA)

I’3.3k = I”3.3k

(mA)

10 8 2.2 4.7 3.3 1.419 0.531 1.95


Vs

(volts)

I

(mA)

I’

(mA)

Vs / I

(kohms)

Vs / I’

(kohms)

10 0.099 0.099 101.01 101.01



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Result:

1. Superposition theorem is verified for the given circuit.

2. Reciprocity Theorem is verified for the given circuit.


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5. Z and Y Parameters

Aim:

To find z & y parameters of a given two port network.

Apparatus:

1. DC voltage source.

2. Resisters. (100Ω, 47Ω, 220Ω, 680Ω, 560Ω).

3. Voltmeter.

4. Ammeter.

5. Breadboard.

Theory:

Networks having two terminals designated as input terminals and two terminals designated as output terminals are

called Two Port Networks. The set of input terminals is called INPUT PORT and the set of output terminals is

called OUTPUT PORT.

A two port network is described by V1, I1, V2, I2 and their inter relations are expressed by

Z parameters normally used in power systems.

Y parameters normally used in power systems.

ABCD parameters used in transmission lines.

H parameters electronics.

Z parameters

V1= Z11I1+Z12I2

V2= Z21I1+Z22I2

Y parameters

I1= Y11V1+Y12V2

I2= Y21V1+Y22V2


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Circuit diagram:

R1

100

R2220

R3

47

R4

680

R5

560

V110 V

V210 V

1 2

3

5

6 0

Pocedure:

Determination of Z parameters

1. Connect a DC voltage source of 5v to the input and measure the current I1. Since I2 =0.

2. Using the same circuit we can determine Z21. For this value we have to measure V2

3. Note that even if a voltmeter is connected at the output there is no current at outputs as voltmeter has a very high

resistance. As I2 =0.

4. To determine Z12, I1must be zero. So do not connect anything at the input.

5. Connect a DC voltage source of 5V at the output and measure the voltage v1 since I1 = 0.

6. Using same circuit we can determine Z22. Since I1 = 0.

Determination of Y parameters

1. To determine Y11, V2 should be zero. So short the output terminals and measure input current

and input voltage. As V2 = 0

2. Using the same circuit we can determine Y21, Measure the short circuit current I2. As V2 =0.

3. To determine Y21, V1 should be zero. So short the terminals through an ammeter


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4. Determine Y22, as V1 = 0, I2 = Y22V2


Z Parameters:

V1= Z11I1+Z12I2

V2= Z21I1+Z22I2

Step1: Open the output terminals.

V310 V

R6

100

R7220

R8

47

R9

560

R10

680

U1

DC 1e-009

0.011 A+ -

4

7

8

9

0

U2DC 10M2.500 V

+

-

11

12

I2=0A

I1=10/ (100+220+560) =0.01136A

V2=0.01136x220=2.42v

Z11=V1/I1=880Ω

Z21=V2/I1=220Ω

Step2: open the input terminals.


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V310 V

R6

100

R7220

R8

47

R9

560

R10

680

9

0

U2DC 10M2.323 V

+

-

4

7

U1DC 1e-0090.011 A

+

-

8

10

11

I1=0A

V2=10V

I2=10/ (47+220+680)=0.0105567A

V1=0.01055x220=2.323V

Z12=V1/I2=220Ω

Z22=V2/I2=947Ω

Z21=Z12

Y Parameters:

I1= Y11V1+Y12V2

I2= Y21V1+Y22V2

Step1: Short the output terminals.

V310 V

R6

100

R7220

R8

47

R9

560

R10

680

9

0

U1DC 1e-0090.012 A

+

-

4

7

8

U2DC 1e-0092.803m A

+

-

10

11

V2=0V

V1=10V


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Req = [(470+680)||220]+660=828.8Ω

I1=10 / (828.8) =0.012A

I2= - (0.012x220)/ (220+727) = -0.0027A

Y11=I1/V1=0.0012mohs

Y21=I2/V1= -0.00027mohs

Step2: Short the input terminals.

V310 V

R6

100

R7220

R8

47

R9

560

R10

680

9

0

U1DC 1e-0090.011 A

+

-

8

10

11

U2

DC 1e-009

2.803m A+-

4

7

V1=0v

V2=10v

Req = [(100+560) ||220] +47+680

=892Ω

I2= (10/892) =0.0112A

I1= - (0.0112x220) / (220+660) = -0.0028A

Y12=I1/V2= -0.00028mohs

Y22=I2/V2=0.000112mohs

Y12=Y21

Observations:

Z parameters:


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V1(v) V2(v) I1(m A) I2(m A) Z11 (k Ω) Z21(k Ω)

10.02 2.48 11.5 0 0.87 0.21

8 2.02 9.2 0 0.87 0.21


2.3 10.02 0 10.6 0.22 0.96

2.75 12 0 12.5 0.22 0.96

Y Parameters

V1(V) V2(V) I1(m A) I2(m A) Y11(m mohs) Y21(m mohs)

10.02 0 12.4 - 2.8 1.2 -0.28

8 0 9.66 - 2.25 1.2 -0.28


0 10.02 -2.8 11.4 -0.28 1.1

0 12 -3.4 13.9 -0.28 1.1


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Using multisim: Z parameters:


10 2.5 11.364 0 0.87 0.21

8 2 9.652 0 0.82 0.21


2.323 10 0 10.56 0.219 0.95

2.78 12 0 12.67 0.219 0.95

Y Parameters

V1(v) V2(v) I1(m A) I2(m A) Y11(m mohs) Y21(m mohs)

10 0 12 2.802 1.2 0.28

8 0 9.652 2.22 1.2 0.28



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0 10 2.803 11 0.28 1.1

0 12 3.363 13 0.28 1.1



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Result:

Z and Y parameters are found for the given 2-port network.


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6. Transmission and Hybrid Parameters.

Aim:

To find out the Transmission and Hybrid parameters of the given two port network.

Apparatus Required:

1. DC Voltage source.

2. Resistors.

3. Voltmeter.

4. Ammeter.

Theory: Networks having two terminals designated as input terminals and two terminals designated as output

terminals are called TWO PORT NETWORKS. The set of input terminals is called INPUT PORT and the set of

output terminals is called OUTPUT PORT.

A two port network is described by V1, I1, V2, I2 and their inter relations are expressed by

Z parameters normally used in power systems.

Y parameters normally used in power systems.

ABCD parameters used in transmission lines.

H parameters used in electronics.

Hybrid parameters:

V1=h11I1+h12V2

I2=h21I1+h22V2

Transmission parameters:

V1=AV2+BI2

I1=CV2+DI2

Procedure:

Hybrid parameters:


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1. Input voltage V1 and output current are taken as dependent variables; these parameters are called Hybrid

Parameters.

2. Keeping the input terminals open I1 = 0 so V1 = h12V2

3. Using the same circuit h22 can be measured, as I1= 0, I2 = h22V2

4. To determine h12 output terminals are shorted through an ammeter as V2 = 0, V1 = h11I1

5. Same circuit can be used to determine h21 also V2 =0, I2 = h21I1


1. A = V1/V2 is measured when receiving end is open circuited.

2. C = I1/V2 is also measured when receiving end is open circuited.

3. B = V1/ I2 is measured when receiving end is shorted.

4. D = I1/I2 is measured when receiving end is shorted.

Circuit Diagram:


Hybrid parameters:


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To determine the h parameters first short circuit output terminal

V1= 10V V2= 0

Req= [47||560] +100

=143.36 ohm

I1= 10/143.36= 0.06975A

I2= - (0.6975x560)/560+47

= -0.06435amps

h11=V1/I1 = 143.369

h21= I2/I1=-0.06435/0.06975= - 0.9225A

Now open input terminals

V1

12 V

R1

100

R2

47

R3560

1

U1DC 10M0.000 V

+

-

4

U2DC 1e-0090.000 A

+

-

2

3

0

V2=12V I1=0


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I2= 12/607 = 0.0197A

v1= 560x0.0197= 11.070V

h12 =V1/V2 = 11.07/12 = 0.922

h22= I2/V2 = 0.0197/12 = 0.00164 mho

h12= - h21


Open the output terminals:

I1= 10/660 = 0.015A

I2= 0A

V1= 10V

V2= 0.015x560

= 8.48V

A= V1/V2 = 10/8.48 = 1.179

C= I1/V2= 0.015/8.48 = 0.00176 mohs

Now short the output terminals


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V1=10V V2=0V

Req = 146.36

I1= 10/143.36 = 0.06995A

I2= (- 0.0699x560)\ (560+47) = - 0.06434A

B= V1/I2 = 10/0.06434 = -155.423ohms

D = I1/I2 = 0.06975/0.06434 = - 1.0841

AD-BC= (1.179x1.0841)-(155.4x0.00176)

=1

Observations:

Hybrid parameters:

V1(v) V2(v) I1(mA) I2(mA) h12 h22( m mho)

11.06 12.02 0 19.28 0.92 1.6

10.15 11.01 0 18.5 0.92 1.6


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V1(v) V2(v) I1(mA) I2(mA)

h11 ohm h21

10.02 0 69.8 -64.5 143.5 -0.92

5.01 0 34.5 -32.0 143.5 -0.92


V1(v) V2(v) I1(mA) I2(mA) A C(m mohs)

10 8.49 15.3 0 1.178 1.8

5.01 4.25 7.6 0 1.178 1.8

V1(v) V2(v) I1(mA) I2(mA) B (mohms) D

10.01 0 69.8 -64.3 -0.155 -1.1

5.01 0 34.5 -32.0 -0.156 -1.1

Using Multisim:

Hybrid parameters:

V1(v) V2(v) I1(mA) I2(mA) h12 h22( m mho)


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11.07 12 0 19.76 0.92 1.6

10.14 11 0 18 0.92 1.6

V1(v) V2(v) I1(mA) I2(mA)

h11 ohm h21

10 0 69.75 -64.35 143.5 -0.92

5 0 35 -32.0 143.5 -0.92


V1(v) V2(v) I1(mA) I2(mA) A C(m mohs)

10 8.48 15.15 0 1.178 1.8

5 4.242 7.576 0 1.178 1.8

V1(v) V2(v) I1(mA) I2(mA) B (mohms) D

10 0 69.75 -64.35 -0.155 -1.1

5 0 35 -32.0 -0.156 -1.1


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Result:

Hybrid parameters, transmission parameters for the given circuit are determined.


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7. Compensation and Milliman’s Theorems.

Aim:

1. Verify compensation theorem for a given network.

2. Verify Milliman’s theorem for a given network.

Apparatus Required:

1. Voltmeter

2. Resistances

3. Bread board

4. Ammeter


Theory:

1) Compensation Theorem:

It states that in any linear bilateral network, any element can be replaced by voltage source of magnitude equal to

current through the element multiplied by value of element provides currents and voltages in another part

of circuit remain unaltered.

Consider the network as shown in figure.


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In the above circuit, the resistance R, can be replaced by a voltage source at value

IR

2) Milliman’s Theorem: Milliman’s theorem states that in any linear active bilateral network consisting of no of voltage sources which are in

parallel and are in series with their internal resistances then this entire system of circuit can be replaced by a single

voltage source in series with a single resistance.

Let us consider the circuit shown below consisting of no of voltage sources V1,V2,V3............Vn are in series with

their internal resistances r1,r2,r3..........rn can be reduced into a single circuit with a voltage source ‘V’ and the

resistance ‘R’ as shown in the figure ‘b’.

Fig (a)


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fig (b)

Circuit Diagram:


R1

3.3kΩ

R2

2.2kΩ

R31kΩ

V112 V

1 3

0

Fig (1)

R1

3.3kΩ

R2

2.2kΩ

R31kΩ

V112 V

1 3

R41.1kΩ

0

2

Fig (2)


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R1

3.3kΩ

R2

2.2kΩ

R31kΩ

3 R41.1kΩ

2

V11.03 V

1

0

Fig (3)

2) Milliman’s Theorem:

V110 V

V212 V

R1100Ω

R2470Ω

R3560Ω23

1

0

Fig (4)

Procedure:


1. Consruct the circuit as shown in figure.

2. Note the ammeter reading I1

3. Modify the circuit in fig (1) as fig (2) and replace R2 with R2+∆R and voltage source

V’=I1-∆R.

4. Note the ammeter reading I2

5. Construct the circuit as in fig (3) and note the ammeter reading I3


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6. Tabulate the above readings by repeating the experiment for 5 sets of resistor values.

2) Miliman’s Theorem:

1. Connect the circuit as shown in fig (a).

2. An ammeter is connected in series with the load resistance R3 and the corresponding load current I1 (IL) is

determined.

3. The circuit is reduced into the equivalent form of thevenins voltage with a resistor of Rth.

4. Now the current across the load is measured as Il’.

5. If the currents Il & Il’ are equal then the milliman’s theorem is verified.

Observations:

1. Compensation theorem:


From fig (1):

R1

3.3kΩ

R21kΩ

R3

2.2kΩ

13

0

Req = (3.3+0.688)*1000

= 3.98 k ohms

I = V/R = 12/3.98k

= 3.009 mA

I1 = I (1/1+2.2)

= 3.009*(1/3.2)

= 0.94 mA


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To find I2 add ∆R =1k ohm

From fig (2):

Req = ((3.3*1/1+3.3)+3.3)*1000

= (0.767+3.3)*1000

= 4.067 k ohms

I = V/R

= 12/4.067*1000

= 2.95 mA

I2 = I (1/1+3.3)

= 2.95 * 0.233

= .686 mA

I’ = I1 – I2

= 0.94 – 0.686

= 0.253 mA

VERIFICATION:

From fig (3):

V = I *∆R

= 0.94*1.1

=1.03 volts

Req = (((2.2+1.1)1/2.2+1.1+1)+3.3)*1000

= (3.3/4.3)+3.3

= 4.067 k ohms

I’ = V/R


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= 1.03/4.067

= 0.253 mA

Both currents are equal

Hence compensation theorem is verified.

Bread board results:

Multisim Results:

V R1 R2 R3 I1

(A)

∆R V’=I1.∆R I2 (A) I3 (A) I1-I2

(A)

12

12

12

3.3k

1k

560

2.2k

3.3k

100

1k

2.2k

100

0.941

2.069m

9.836m

1.1k

1.1k

470

0.99

2.276

4.622

0.686

1.622m

2.776m

0.255

0.446

7.057

0.255

0.447

7.060

12

12

100

2.2k

560

1k

470

560

0.015

1.683m

100

100

1.5

0.168

13m

1.575m

2.02m

0.109

2m

0.108

V R1 R2 R3 I1

(A)

∆R V’=I1.∆R I2 (A) I3 (A) I1-I2

(A)


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2. Millimans theorem:


From fig (4):

Applying K.C.L,

(V-10)/100 + (V-12)/470 +V/560 = 0

V(1/100 + 1/470 + 1/560) = (10/100 +12/470)

V(0.01 + 0.002 + 0.002) = 0.1+0.026

V(0.014) = 0.126

V = 9 volts

I = V/R

= 9/560

= 0.016A

12

12

12

3.3k

1k

560

2.2k

3.3k

100

1k

2.2k

100

0.941

2.069m

9.836m

1.1k

1.1k

470

0.99

2.276

4.622

0.686

1.622m

2.776m

0.255

0.446

7.057

0.255

0.447

7.060

12

12

100

2.2k

560

1k

470

560

0.015

1.683m

100

100

1.5

0.168

13m

1.575m

2.02m

0.109

2m

0.108


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Using millimans theorem,

R’=1/ (G1+G2)

= 1/ ((1/100) + (1/470))

= 1/ (0.01+0.002)

= 83.3 ohms

V’= (V1G1 +V2G2)/ (G1+G2)

= ((10/100) + (12/470))/ (0.012)

= 0.126/0.012

= 10.5 volts

R283.3Ω

R3560Ω

V110.5 V

1

2

3 I’ = V’/Req

= 10.5/ (83.3+560)

= 0.016A

Both currents are equal,

Hence millimans theorem is verified.

Breadboard results:


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V1

(V)

V2

(V)

R1 R2 R3 I

(mA)

I’

(mA)

V’

(V)

R’

10 12 100 470 560 16.2 16.1 10.5 82.46

Multisim results:

V1

(V)

V2

(V)

R1 R2 R3 I

(A)

I’

(A)

V’

(V)

R’

10 12 100 470 560 0.016 0.016 10.5 82.46



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Result:

1. Compensation Theorem is verified.

2. Milliman’s Theorem is verified.


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8. Series Resonance

Aim:

1. To observe the resonance and calculate resonant frequency, band width, quality factor in series resonance

circuit.

Apparatus Required:

1. AC voltage source.

2. Resistor.

3. Inductor.

4. Capacitor.

5. Voltmeter

Theory:

Series Resonance:

As the frequency is varied in a RLC circuit maximum current is observed at a particular frequency. This

phenomenon is called series resonance. Also referred to as current resonance. Z

Circuit Diagram:

A) Series Resonance:

R1

100

L1

10mH V1

4 V

50 Hz

0Deg C1 .100u

U1

DC 10M

0.000 V + -

U2

DC 10M

0.000 V + -

U3 DC 10M 0.000 V

+

-

1 2 3

0


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Procedure:

A) Series Resonance:

1. Connect resistor, inductor and capacitor in series.

2. Using the formula.

ƒr=1/ (2П√LC)

Calculate resonant frequency.

3. Note down current through the circuit, Voltage across (VR ), Voltage across Inductor (VL),Voltage across

capacitor (Vc)

4. Plot the graph Current Vs Frequency and Impedance Z Vs Frequency.

5. Plot the graph VRVs Frequency, VLVs Frequency and VCVs frequency.

6. From the graph note down the frequency at which Vc is maximum (Fc), the frequency at which Vr is maximum

(Fr) and the frequency at which Vl is maximum (Fl).

It is observed that Vc becomes maximum at a frequency lower than the resonant frequency and Vl becomes

maximum at a frequency more than the resonant frequency.

7. Frequency at which Vc becomes maximum can be calculated using the formula.

ƒc=1/2П ((1/LC)-(R*R/2L)) 1/2

Frequency at which Vl becomes maximum can be calculated using the formula.

ƒl=1/2П ((1/LC)-(R*R*C*C)/2)1/2

Verify with observed values.

8. On the graph current Vs frequency, note down the maximum current.Calaculate 70.7% of this current and draw a

horizontal line corresponding to this value on the graph. Note down the values at which this horizontal line

intersects the curve (f1 and f2).

9. The average of frequencies f2-f1 is called Band Width (BW).

10. fr/BW is known as Q (quality factor).


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Calculate Q using Q=BW/fr and also Q=Xlr/R

=2Пfrl/R Where Xlr is reactance of inductor at resonant frequency.

11. Voltage across capacitor =IXc=V/ώrCR=VώrL/R=QV.

Calculate the ratio of voltage across Capacitor to applied voltage. Observe that ratio (amplification) is

=Q.High Q coils are sometimes used to produce high voltages.


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OBSERVATION:-

S.NO: FREQUENCY

(hz)

APPLIED

VOLTAGE

Va(volts)

Vr (volts) VL (volts) Vc (volts) CURRENT (I)

(amps)

1 50 4 3.8 0.12 1.21 0.038

2 100 4 3.96 0.24 0.65 0.0389

3 150 4 4 0.38 0.43 0.039

4 159.2 4 4 0.4 0.396 0.04

Practical

Values

Multisim

Values

1

2

3

4


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GRAPHS:-

CURRENT~FREQUENCY :-


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CAPACITOR VOLTAGE~FREQUENCY:-

INDUCTOR VOLTAGE~FREQUENCY:-


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RESULT:-

1. Resonant frequency=159.2 Hz

2. Band Width=1575 Hz

3. Quality Factor=0.101


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9. Parallel Resonance

Aim:

To observe the resonance and calculate resonant frequency, band width, quality factor in parallel resonance

circuit

Apparatus Required:

1. AC voltage source.

2. Resistor.

3. Inductor.

4. Capacitor.

5. Voltmeter

Theory:

Parallel Resonance:

As the frequency is varied in a RLC circuit maximum voltage is observed at a particular frequency. This

phenomenon is called Parallel resonance. Also referred to as voltage resonance.

Circuit Diagram:

Parallel Resonance


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R110kΩ

R2

100Ω

L110mH

C110uF

XMM1

XMM2

2

V2

10 Vrms

50 Hz

0°

1

3

Procedure:

Parallel Resonance:

1. Connect a voltmeter across the parallel combination and note down voltage as frequency is gradually increased.

You will note that voltage will be maximum at a certain frequency. This frequency is known as resonant frequency.

Note down the voltage across series resistor.

2. Note down the maximum value of voltage and mark a horizontal line at 0.707 times Vmax. At the points of

intersection mark f1 & f2 known as half power frequencies.

S.NO FREQUENCY

(HZ)

APPLIED

VOLTAGE

(Va)volts

Vr

(volts)

Vout

(volts)

I=Vr/r

(AMPS)

Z=V/I

(ohms)

1

2

3

4

5

503

550

450

350

300

10

10

10

10

10

1.515

4.52

5.315

8.963

9.467

8.485

7.657

7.271

3.806

2.764

0.01515

0.0452

0.05315

0.08963

0.09467

560.066

169.402

136.8

42.463

29.196


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CALCULATIONS:

Fr = (1/ (2П√lc))

=1/ (2*∏√10*10*10ˆ-9)

Fr =503.292 hz

Xl=j31.16

Xc=-j31.162….

3. Draw the curves

Vout Vs frequency

I Vs frequency

Z Vs frequency.

4. Calculate Half power frequencies f1 and f2 using the formula.

ω1=-1/2RC+ [(1/2RC) 2

+1/LC] 1/2

(Lower Half power Frequency)

ω2=1/2RC+ [(1/2RC) 2

+1/LC] 1/2

(Upper Half power Frequency).

5. Band Width = ω2- ω1=1/RC.

f2-f1=1/2ПRC.

Quality Factor=ωrRC.

Observations:

S.NO FREQUENCY

(HZ)

APPLIED

VOLTAGE

(Va)volts

Vr

(volts)

Vout

(volts)

I=Vr/r

(AMPS)

Z=V/I

(ohms)


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1

2

3

4

5


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RESULT:

Parallel Resonance is verified


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10. LOCUS OF CURRENT VECTOR IN AN R-L CIRCUIT

CONTENT:

In this experiment you will learn that current vector leads the applied voltage and

the tip of the current vector describes a semi circle when one of the components (R or L) is

varied from zero to infinity.

CIRCUIT 1:An RL circuit is shown below:

v

current I =--------

R+jXL

V(R+JXL) V.R jVXL

-------------------- = ---------- + -------------

R2+XL

2 R

2+XL

2 R

2+XL

2

Z=( R2+XL

2)

1/2


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VR jVXL

SO I= ----- + --------- =Ix+jIy say

Z2 Z

2

Two cases arises:

a) Keep XL constant and vary R (different resistors used)

b) Keep R constant and vary Xl (different inductors used ) In either case tip of the current vector describes a semi circle.

PROCEDURE:

CIRCUIT 1:


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These are three methods to draw the locus

METHOD 1:

Using a multimeter in AC voltage range, note down voltage applied, voltage

across resistor and voltage across capacitor. Keeping resistor constant and for various values of

capacitor, note down meter readings and fill up the following table:

Keeping C constant, use values of R and note down Vapplied, Vr and VL producing a table

similar to above.

S.No Vapplied Vr VL

1

2

3

4

5


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For each set of readings a triangle can be constructed using a compass as shown. All the points

such as A, B etc., lie on a semi circle.

METHOD 2:

Connect oscilloscope channel 1 and channel 2 as shown in circuit 1.

The wave forms are as shown below when the oscilloscope is kept in dual mode.


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The time delay between the two waves is t. T is the period. Both t and T are noted for each value

of capacitor. The angle between the two wave forms is

a = t*360/T

Measure also magnitudes Va and Vl from the oscilloscope. Draw the triangle as shown. Different

triangles can be constructed for different values of capacitor. Tips of all such triangles fall on a

semi circle.

METHOD 3:


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Third method uses Lissazous figure to measure angle between the applied voltage

and voltage across resistor Vr.

By making connections as above, channel 1 displays the applied voltage and channel 2 displays

the voltage across the resistance.

If we select dual-trace option Va and Vr are displayed simultaneously and the time lag between

the two can be measured and converted to angle.

If we select XY option we can display the Lissazous figure and angle can be obtained using the

formula

Y1 X1

Sina = ----- = ------

Y2 X2


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‘AB’ is marked proportional to applied voltage. Using a protractor mark a line at an angle mark a

to AB. Mark the magnitude of voltage across resistance on this line to get the point P, join PB.

Using several values of resistor R repeat the experiment. It can be observed that for each resistor

value a different location for point P is obtained. It is also observed that all the points P1, P2,

P3…. Fall on a semi circle.


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Take reading as shown below

S.NO Vr Vs X1 X2 Y1 Y2 sina=Y1/Y2 a

1 4 2.2 2 2.2 3.6 4 0.9089 65.39

2 4 2 1.8 2 3.6 4 0.8995 64.1

3 4 1.6 1.6 1.6 3.6 4 1 90

4 4 0.6 1 1.2 3.8 4 0.8332 56.43

NOTE:

All the above three methods can be used to obtain the locus of current vector in the case where

capacitor value is kept unchanged and various values of resistors are used.

Also note that the current vector in this experiment is actually represented by voltage across the

resistor to scale.


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11. LOCUS OF CURRENT VECTOR IN AN R-C CIRCUIT

CONTENT:

In this experiment you will learn that current vector leads the applied voltage and

the tip of the current vector describes a semi circle when one of the components (R or C) is

varied from zero to infinity.

CIRCUIT 1:

An RC circuit is shown below:

V

current I =--------

R+jXc

V(R+JXc) V.R jVXc

-------------------- = ---------- + -------------

R2+Xc

2 R

2+Xc

2 R

2+Xc

2

Z=( R2+Xc

2)

1/2


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VR jVXc

SO I= ----- + --------- =Ix+jIy say

Z2 Z

2

Two cases arise:

a) Keep Xc constant and vary R (different resistors used)

b) Keep R constant and vary Xl (different capacitors used ) In either case tip of the current vector describes a semi circle.

PROCEDURE:

CIRCUIT 1:


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These are three methods to draw the locus

METHOD 1:

Using a multimeter in AC voltage range, note down voltage applied, voltage

across resistor and voltage across capacitor. Keeping resistor constant and for various values of

capacitor, note down meter readings and fill up the following table:

S.No Vapplied Vr Vc

1

2

3

4

5

Keeping C constant, use values of R and note down Vapplied, Vr and Vc producing a table

similar to above.

For each set of readings a triangle can be constructed using a compass as shown. All the points

such as A, B etc., lie on a semi circle.


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METHOD 2:

Connect oscilloscope channel 1 and channel 2 as shown in circuit 1.

The wave forms are as shown below when the oscilloscope is kept in dual mode.

The time delay between the two waves is t. T is the period. Both t and T are noted for each value

of capacitor. The angle between the two wave forms is

a = t*360/T


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Measure also magnitudes Va and Vl from the oscilloscope. Draw the triangle as shown. Different

triangles can be constructed for different values of capacitor. Tips of all such triangles fall on a

semi circle.

METHOD 3: Third method uses Lissazous figure to measure angle between the applied

voltage and voltage across resistor Vr.

By making connections as above, channel 1 displays the applied voltage and channel 2 displays

the voltage across the resistance.

If we select dual-trace option Va and Vr are displayed simultaneously and the time lag between

the two can be measured and converted to angle.

If we select XY option we can display the Lissazous figure and angle can be obtained using the

formula

Y1 X1

Sina = ----- = ------

Y2 X2


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‘AB’ is marked proportional to applied voltage. Using a protractor mark a line at an angle mark a

to AB. Mark the magnitude of voltage across resistance on this line to get the point P, join PB.

Using several values of resistor R repeat the experiment. It can be observed that for each resistor

value a different location for point P is obtained. It is also observed that all the points P1, P2,

P3…. Fall on a semi circle.


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Take reading as shown below

S.NO Vr Vs X1 X2 Y1 Y2 sina=Y1/Y2 A

1 4 2.2 2 2.2 3.6 4 0.9089 65.39

2 4 2 1.8 2 3.6 4 0.8995 64.1

3 4 1.6 1.6 1.6 3.6 4 1 90

4 4 0.6 1 1.2 3.8 4 0.8332 56.43

NOTE:

All the above three methods can be used to obtain the locus of current vector in the case where

capacitor value is kept unchanged and various values of resistors are used.

Also note that the current vector in this experiment is actually represented by voltage across the

resistor to scale.


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12.Measurement of 3-phase power by two wattmeter method for

unbalanced loads.

Objective:

Measurement of power by 2-wattmeters for unbalanced loads in a 3- phase circuit.

Apparatus:

32 Amps, 3 pole Fuse Switch

0 -300 W, U.P.F. Wattmeter’s

0 – 10 A, Ammeter

0-300 V, Voltmeter

Theory:

In a 3-phase, 3-wire system, power can be measured using two wattmeter’s for balance and

unbalanced loads and also for star, delta type loads. This can be verified by measuring the power

consumed in each phase. In this circuit, the pressures coils are connected between two phase

such that one of the line is coinciding for both the meters.

P1 + P2 = 3 VPh IPh COSø

Power factor Cosø = Cos (tan-1

√3 ((P1 –P2)/ (P1 +P2)))

Circuit diagram:


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Observations:

Type of Load

(W)

W1

W

W2

W

I1

Ma

I2

mA

Vph

Volts

W1+

W2

KW

P

KW

R1=R2=R3=1K

R1+L1=1K+40m

R1+C1=1K+1uf

52.908

52.904

4.757

52.908

52.904

4.757

230.018

229.99

68.9

230.018

229.99

68.94


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FOR UNNBALANCED LOADS:

TYPE OF LOAD

W1(W)

W2(W)

I1

(mA

I2

(mA

Vph

(V)

W1+W2

(W)

P

(KW)

R1=560,

R2=1K,

R3=220

R1+L1=560+1m,

R2+L2=1K+10m

R3+L3=220+20m

R1+C1=560+1uf

R2+C2=1K+1uf

R3+C3=220+10uf

2.832

94.746

2.832

77.802

240.393

77.802

594.456

410.746

594.456

71.169

1.045

71.169

TYPE OF LOAD

W1(W)

W2(W)

I1

(mA

I2

(mA

Vph

(V)

W1+W2

(W)

P

(KW)


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Result:

Three Phase Power Measured by two wattmeter method for unbalanced load is

13. Measurement of Active and Reactive power by star and delta connected

balanced loads.

Objective:

Measurement of active and reactive power using 1-wattmeter at different R, L & C loads.

Apparatus:

Hardware: Name of the apparatus Quantity

32 Amps, 3 pole Fuse Switch 1 No

0 -300 W, U.P.F. Wattmeters 1 No

0 – 10 A, A.C Ammeter 1 No

0-300 V, A.C Voltmeter 1 No


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Theory:

The active power is obtained by taking the integration of function between a particular time intervals

from t1 to t2

t2

P = 1/ (t2- t1) P (t) dt

t1

By integrating the instantaneous power over one cycle, we get average power.

The average power dissipated is

Pav = Veff[ Ieff cosθ]

From impedance triangle,

Cosθ = R/Z

Substituting, we get

Reactive Power Pr = Veff[ Ieff sinθ]

Active power measurement:


GRIET/EEE 117

Reactive power measurement:


GRIET/EEE 118

Procedure:

a) Connect the circuit as shown in the circuit diagram.

b) Keep all the toggle switches in ON condition.

c) Switch on equal loads on each phase i.e. balanced load must be maintained with different

load combinations.

d) Connect the ammeter in R-Phase and then switch OFF the toggle switch connected across

the ammeter symbol.

e) Connect the pressure coil of the wattmeter across R-Y phase and current coil in R-phase

to measure active power.

Observations:

Load: Balanced load

Active Power:


GRIET/EEE 119

Type of load Vph

(Volts)

Il

(mA)

Pph

(Watts)

Pactual

P=3*Pph

(Watts)

Cosθ

=P/( 3VlIl )

R=10k 120.009 11.992 1.439 4.296 0.986

R-10k

C=1µF

120.009 11.432 1.302 3.906 0.949

L=1mH

F

120.009 100 12 36 0.999

Reactive Power:

Type of the

load

Vph

(Volts)

Il

(mA)

Pph

(Var)

Pactual

P=3*Pph

(Var)

Cosθ

=P/( 3VlIl )

R=10k 120.009 11.992 4.150 12.45 0.9612

R-10k

C=1µF

120.009 11.432 1.602 4.806 0.389

L=1Mh

F

120.009 99.647 9.4 28.287 0.788


GRIET/EEE 120

Result: Active and Reactive powers were measured using 1-wattmeter at R, L and C Loads.


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Multisim/Networks Lab

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