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
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:
GRIET/EEE
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:
2.2k
R1
R2
1
2.2k
0
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:
2.2k
2
R1
1
R2
2.2k
V1
10 V
00
I=10/4.4*103 =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:
1.1k
R1
2
V1
5.006 V
0
1
Finding current through 1k ohm resistor using Thevenin’s theorem,
1.1k
2
R1
R2
3
1k
V1
5.006 V
0
0
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
5. DC voltage source
Theory:
Norton’s Theorem:
Any linear circuit containing several energy sources and resistance can be replaced by a single constant generator
parallel with a single resistor.
Circuit diagram:
1
R1
R2
2
100
V1
10 V
150
R3
51
3
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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 equ
Rn (Norton’s resistance)
4. Construct an equivalent circuit and verify the current across the load in both circuits.
Theoretical calculations:
1
R1
R2
2
100
150
V1
10 V
R3
51
3
STEP 1:
Finding R equivalent:
To find R
R1
1
100
R2
51
0
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Req= (100*51)/151
=33.77ohm
STEP 2:
To find IN:
R1
1
4
100
V1
10 V
R2
51
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
1
I1
0.1 A
R1
33.77
150
R2
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) =
Theoretical Calculations to be done by Students:
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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
5. DC voltage source
Theory:
Maximum Power transfer theorem:
Maximum power transfer theorem states that the power delivered from a source to a load is maximum when sour
resistance equals load resistance.
Circuit Diagram:
1
R1
RL
a
2.2k
V1
10 V
R3
2.2k
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
R(load)Ώ
V2/4RL
To load V×I
1.
2.381
2.381
5.66m
1k
2.
2.5
2.273
5.68m
1.1k
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
5.68
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.
R2
3
2.2k
R4
2.2k
0
Rab= ((2.2K)//(2.2k))/94.4k)
=1.1k Ώ
STEP 2:
Finding VTH
1. VTH is the voltage across a&b.
R5
4
2.2k
2
V2
10 V
R6
2.2k
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:
1
R1
2.2k
V1
10 V
R2
2
10
4.7k
R3
3.3k
V2
8V
0
.
B) Reciprocity Theorem:
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R2
6
+
-
R4
2
2.2k
0.099m
A
3
10k
U1
DC 1e-009
R5
4.7k
R1
3.3k
R3
1.0k
4
5
V1
10 V
0
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
V2
R1
R2
R3
I1
I2
I’3.3k
(volts)
(volts)
(ohms)
(ohms)
(ohms)
(mA)
(mA)
=
I’’3.3k
(mA)
10
2
2.2k
4.7k
3.3k
1.4194
0.5316
1.951
2) Reciprocity Theorem Experiment:
R1
R2
R3
R4
R5
Vs
I
I’
Vs/I
Vs/I’
(kohms) (kohms) (kohms) (kohms)(kohms) (volts) (mA) (mA)
(kohms) (kohms)
3.3
111.11 111.111
2.2
1
10
4.7
10
0.09
0.09
Theoretical calculations:
1) Superposition theorem:
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R1
1
R2
2.2k
10
4.7k
2
R3
3.3k
V1
10 V
V2
8V
0
Consider 10V D.C. voltage source and replace 8V D.C. voltage source with short circuit.
1
R1
R2
2
2.2k
4.7k
+
-
1.419m
V1
10 V
U2
A DC 1e-009
4
R3
3.3k
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
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= 1.4194 mA.
Now, consider 8V D.C. voltage source and replace 10V D.C. voltage source by short circuit.
8
5
4.7k
+
-
R6
U1
A DC 1e-009
0.531m
R4
2.2k
V2
8V
7
R5
3.3k
0
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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3
R7
2.2k
R9
6
4.7k
+
-
V3
10 V
9
1.950m
U3
A DC 1e-009
V4
8V
11
R8
3.3k
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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6
R2
R4
2
2.2k
+
3
10k 
U1
0.099m
-
A DC 1e-009 
R5
4.7k 
R1
3.3k
R3
1.0k
4
5
V1
10 V
0
Applying nodal analysis,
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’
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R2
R4
2
2.2k
3
10k
1
R1
3.3k
+
R3
1.0k
-
0.099m
6
A
U1
DC 1e-009
4
R5
4.7k
V1
10 V
0
Applying nodal analysis,
(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:
1) Superposition theorem:
V1
V2
R1
R2
R3
I1
I2
I’3.3k
(volts)
(volts)
(kohms)
(kohms)
(kohms)
(mA)
(mA)
=
I”3.3k
(mA)
10
8
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2.2k
4.7k
3.3k
1.42
0.5
1.92
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2) Reciprocity theorem:
Vs
I
I’
Vs / I
Vs / I’
(volts)
(mA)
(mA)
(kohms)
(kohms)
10
0.1
0.1
100
100
Multisim Results:
1) Superposition theorem:
V1
V2
(volts) (volts)
10
8
I2
I’3.3k = I”3.3k
(kohms) (kohms) (kohms) (mA)
(mA)
(mA)
2.2
0.531
1.95
R1
R2
4.7
R3
3.3
I1
1.419
2) Reciprocity theorem:
Vs
I
I’
Vs / I
Vs / I’
(volts)
(mA)
(mA)
(kohms)
(kohms)
10
0.099
0.099
101.01
101.01
Theoretical Calculations to be done by Students:
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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 ar
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
5
R3
1
100
2
47
R2
220
V1
10 V
6
R5
560
0
V2
10 V
R4
3
680
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 hi
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
Theoretical calculations:
Z Parameters:
V1= Z11I1+Z12I2
V2= Z21I1+Z22I2
Step1: Open the output terminals.
U1
A
+
0.011
4
R6
R8
9
100
11
47
DC 1e-009
7
R7
220
V3
10 V
+
-
R9
8
0
560
R10
2.500
V
U2
DC 10M
12
680
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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R6
4
R8
9
100
10
47
+
+
-
V
2.323
R7
220
U2
DC 10M
7
0.011
A
U1
DC 1e-009
8
V3
10 V
R9
R10
0
560
11
680
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.
4
R6
R8
9
100
+
47
10
U1
A DC 1e-009
0.012
+
R7
220
7
-
V3
10 V
2.803m
A
U2
DC 1e-009
11
R9
8
0
560
R10
680
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.
U2
-
2.803m
+
A
4
R6
R8
9
100
10
47
DC 1e-009
+
R7
220
7
0.011
A
U1
DC 1e-009
8
V3
10 V
R9
0
560
R10
11
680
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
V1(v)
V2(v)
I1(m A)
I2(m A)
Z12 (k Ω)
Z22(k Ω)
2.3
10.02
0
10.6
0.22
0.96
2.75
12
0
12.5
0.22
0.96
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
V1(V)
V2(V)
I1(m A)
I2(m A)
Y12(m mohs)
Y22(m mohs)
0
10.02
-2.8
11.4
-0.28
1.1
0
12
-3.4
13.9
-0.28
1.1
Y Parameters
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Using multisim: Z parameters:
V1(v)
V2(v)
I1(m A)
I2(m A)
Z11 (k Ω)
Z21(k Ω)
10
2.5
11.364
0
0.87
0.21
8
2
9.652
0
0.82
0.21
V1(v)
V2(v)
I1(m A)
I2(m A)
Z12 (k Ω)
Z22(k Ω)
2.323
10
0
10.56
0.219
0.95
2.78
12
0
12.67
0.219
0.95
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
V1(V)
V2(V)
I1(m A)
I2(m A)
Y12(m mohs)
Y22(m mohs)
Y Parameters
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0
10
2.803
11
0.28
1.1
0
12
3.363
13
0.28
1.1
Theoretical Calculations to be done by Students:
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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
Transmission parameters:
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:
Theoretical Calculations:
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
R1
4
100
+
-
0.000
V
3
R2
1
-
47
U1
DC 10M
+
U2
DC 1e-009
2
R3
560
12 V
0
V2=12V
A
0.000
V1
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
Transmission parameters:
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
Transmission parameters:
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
I1(mA)
I2(mA)
h12
h22( m mho)
Using Multisim:
Hybrid parameters:
V1(v)
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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
Transmission parameters:
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
5. DC voltage source
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
parallel and are in series with their internal resistances then this entire system of circuit can be replaced by a sing
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 w
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:
1) Compensation Theorem:
R1
1
R2
3.3kΩ
2.2kΩ
3
V1
12 V
R3
1kΩ
0
Fig (1)
R1
1
3.3kΩ
V1
12 V
R2
3
2
2.2kΩ
R4
1.1kΩ
R3
1kΩ
0
Fig (2)
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R1
R2
3.3kΩ
2
2.2kΩ
3
R4
1.1kΩ
1
R3
1kΩ
V1
1.03 V
0
Fig (3)
2) Milliman’s Theorem:
1
V1
10 V
V2
12 V
3
2
R1
100Ω
R3
560Ω
R2
470Ω
0
Fig (4)
Procedure:
1) Compensation Theorem:
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 I 3
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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:
Theoretical Calculations:
From fig (1):
3
R1
R3
1
3.3kΩ
2.2kΩ
R2
1kΩ
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:
V
R1
R2
R3
I1
∆R
V’=I1.∆R
I2 (A)
I3 (A)
I1-I2
(A)
(A)
12
3.3k
2.2k
1k
0.941
1.1k
0.99
0.686
0.255
0.255
12
1k
3.3k
2.2k
2.069m
1.1k
2.276
1.622m
0.446
0.447
12
560
100
100
9.836m
470
4.622
2.776m
7.057
7.060
12
100
560
470
0.015
100
1.5
13m
2.02m
2m
12
2.2k
1k
560
1.683m
100
0.168
1.575m
0.109
0.108
Multisim Results:
V
R1
R2
R3
I1
∆R
V’=I1.∆R
I2 (A)
I3 (A)
I1-I2
(A)
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12
3.3k
2.2k
1k
0.941
1.1k
0.99
0.686
0.255
0.255
12
1k
3.3k
2.2k
2.069m
1.1k
2.276
1.622m
0.446
0.447
12
560
100
100
9.836m
470
4.622
2.776m
7.057
7.060
12
100
560
470
0.015
100
1.5
13m
2.02m
2m
12
2.2k
1k
560
1.683m
100
0.168
1.575m
0.109
0.108
2. Millimans theorem:
Theoretical calculations:
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
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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
2
R2
83.3Ω
1
R3
560Ω
V1
10.5 V
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
V2
(V)
(V)
10
12
R1
100
R2
R3
470
560
R2
R3
I’
V’
(mA)
(mA)
(V)
16.2
16.1
10.5
82.46
I
I’
V’
R’
(A)
(A)
(V)
0.016
0.016
10.5
I
R’
Multisim results:
V1
V2
(V)
(V)
10
12
R1
100
470
560
82.46
Theoretical Calculations to be done by Students:
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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:
U1
+
0.000
DC 10M
R1 
1
V1
100 
U2
V
+
0.000
2
DC 10M
L1
V
3
10mH
+
4V
50 Hz
0Deg
C1 .100u
0.000
V
U3
DC 10M
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 (V R ), Voltage across Inductor (VL),Voltage acr
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
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
Vr (volts)
VL (volts)
Vc (volts)
CURRENT (I)
(amps)
Va(volts)
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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XMM1
R2
1
100Ω
2
R1
10kΩ
V2
10 Vrms
50 Hz
0°
L1
10mH
C1
10uF
XMM2
3
Procedure:
Parallel Resonance:
1. Connect a voltmeter across the parallel combination and note down voltage as frequency is gradually increas
You will note that voltage will be maximum at a certain frequency. This frequency is known as resonant frequen
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
intersection mark f1 & f2 known as half power frequencies.
S.NO
FREQUENCYAPPLIED
(HZ)
Vr
VOLTAGE (volts)
Vout
(volts)
(Va)volts
I=Vr/r
Z=V/I
(AMPS)
(ohms)
1
503
10
1.515
8.485
0.01515
560.066
2
550
10
4.52
7.657
0.0452
169.402
3
450
10
5.315
7.271
0.05315
136.8
4
350
10
8.963
3.806
0.08963
42.463
5
300
10
9.467
2.764
0.09467
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
FREQUENCYAPPLIED
(HZ)
VOLTAGE (volts)
(Va)volts
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Vout
(volts)
I=Vr/r
Z=V/I
(AMPS)
(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
-------------------- = ---------R2+XL2
R2+XL2
jVXL
+ ------------R2+XL2
Z=( R2+XL2)1/2
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VR
jVXL
SO I= ----- + --------Z2
=Ix+jIy say
Z2
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:
S.No
Vapplied
Vr
VL
1
2
3
4
5
Keeping C constant, use values of R and note down Vapplied, Vr and VL producing a table
similar to above.
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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
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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+Xc2
R2+Xc2
R2+Xc2
Z=( R2+Xc2)1/2
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VR
jVXc
SO I= ----- + --------Z2
=Ix+jIy say
Z2
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
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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
W2
I1
I2
Vph
W
W
Ma
mA
Volts
W1+
W2
P
KW
KW
R1=R2=R3=1K
52.908
52.908
230.018
230.018
R1+L1=1K+40m
52.904
52.904
229.99
229.99
R1+C1=1K+1uf
4.757
4.757
68.9
68.94
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FOR UNNBALANCED LOADS:
TYPE OF LOAD
I1
I2
Vph
W1+W2
P
(V)
(W)
(KW)
W1(W)
W2(W)
(mA
(mA
2.832
77.802
594.456
71.169
94.746
240.393
410.746
1.045
2.832
77.802
594.456
71.169
I1
I2
Vph
W1+W2
P
(mA
(mA
(V)
(W)
(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
TYPE OF LOAD
W1(W)
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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
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1 No
0 – 10 A, A.C Ammeter
1 No
0-300 V, A.C Voltmeter
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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:
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Reactive power measurement:
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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:
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Type of load
Vph
Il
Pph
Pactual
Cosθ
(Volts)
(mA)
(Watts)
P=3*Pph
=P/( 3VlIl )
(Watts)
R=10k
120.009
11.992
1.439
4.296
0.986
R-10k
120.009
11.432
1.302
3.906
0.949
120.009
100
12
36
0.999
C=1µF
L=1mH
F
Reactive Power:
Type of the
load
Vph
Il
Pph
Pactual
Cosθ
(Volts)
(mA)
(Var)
P=3*Pph
=P/( 3VlIl )
(Var)
R=10k
120.009
11.992
4.150
12.45
0.9612
R-10k
120.009
11.432
1.602
4.806
0.389
120.009
99.647
9.4
28.287
0.788
C=1µF
L=1Mh
F
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Result:
Active and Reactive powers were measured using 1-wattmeter at R, L and C Loads.
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