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LABORATORY MANUAL
BASIC ELECTRICAL ENGINEERING
B-Tech . II Year-Sem-II-ECE
Prepared By:-
P.N.Ramulu Lab - In charge
DEPARTEMENT
OF
ELECTRICAL AND ELECTRONICS ENGINEERING
SIDDHARTHA INSTITUTE OF ENGINEERING &TECHNOLOGY
IBRAHIM PATNAM - R.R. DIST. 501 506
***********
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1
Ex.No.01
Verification of Kirchhoff ’s Current Law
AIM: To verify the Kirchhoff‟s Current Law for a given circuit.
Theory:-
This law states that the algebraic sum of currents meeting at a junction of conductors is zero. In
other words , the sum of currents flowing away from a junction is equal to the sum of currents
flowing towards the junction.
Kirchhoff‟s current law is nothing more than a restatement of principle of conservation of
charge.
Since the amount of charge entering a junction at an instant must be same as the amount of
charge leaving the junction.
Apparatus required:-
S.No Equipment Range Quantity
1 DC.RPS-Voltage Source 0-30 Volts/2A 1
2 Resistors 1.1KΩ,2.2KΩ,3.3KΩ 3
3 Ammeter-DC 0-200 milliamps 3
4 Connecting wires Single lead As required
Given Circuit
PROCEDURE:-
1. Connect the circuit as per the figure shown above
2. Adjust the V1 voltage as 10 volts, and switch ON the supply.
3. Measure the current flowing through R1, R2 , R3 resistors using Ammeters ie, IT, I1 & I2.
4. Tabulate the readings in the tabular column
5. Verify that the I total. = I1 + I2
6. Repeat the procedure for different voltage values, and then switch OFF the supply.
7. Compare the values Practical to Theoretical.
Tabular column:
S.No I1
(mA)
I2
(mA)
IT
(mA)
I1 + I2
(mA)
1
2
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Theoretical Calculations:-
Case1. (For measuring I1)When the V1 source is applied
REq = (R2 ║ R3)+R1
IEq = V1 / REq
I1 = IEq X R2
R2+R3
Case 2. (For measuring I2)When the V1 source is applied
REq = (R2 ║ R3)+R1
IEq = V1 / REq
I2 = IEq X R3
R2+R3
Comparison of values Practical with Theoretical
KCL Theorem Case-1 (I1 ) Amp Case-2 (I2) Amp Case-3 ( I-total) Amp
Theoretical Values
Practical Values
Safety Precautions:
1. Reading must be taken without parallax error.
2. Measuring instruments must be connected properly & should be free from errors.
3. All connections should be free from loose contacts .
4. The direction of currents should be identified correctly
Result:-
Kirchhoff‟s Current Law is verified practically.
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Ex.No.02
Verification of Kirchhoff ’s Voltage Law
AIM: To verify the Kirchhoff‟s Voltage Law for a given circuit.
Theory:-
This law states that any time instant the algebraic sum of voltages around a closed circuit or a
loop is zero. That is , for a closed circuit having “k” elements ,
K
∑Vj=0
J=0
Where Vj represents the voltage drop of the j th element .
V1+V2+V3+……….Vk = 0
This statement simply tells us that if we start from a particular junction and go around a closed
circuit so as to come back to the same junction , the net potential drop (or potential rise ) is
zero, Because we have come back to the point at the same potential .
Kirchhoff’s Voltage Law can also be stated as : in any closed circuit the algebraic sum of the
products of current and resistance in each of the conductors is equal to the algebraic sum of
the emf‟s of the batteries
Apparatus required:-
S.No Equipment Range Quantity
1 DC.RPS-Voltage Source 0-30 Volts/2A 1
2 Resistors 1.1KΩ, 2.2KΩ 2
3 Ammeter-DC 0-200 milliamps 1
4 Voltmeter - DC 0-30V 2
5 Connecting wires Single lead As required
Given Circuit
PROCEDURE:-
1. Connect the circuit as per the figure shown above
2. Adjust the V voltage as 10 volts, and switch ON the supply.
3. Note the reading of Ammeter & voltmeter ,ie, I , and V1, V2 ,V3 from the meters.
4. Tabulate the readings in the tabular column
5. Verify that the V total. = V1 + V2
6. Repeat the procedure for different voltage values, and then switch OFF the supply.
7. Compare the values Practical to Theoretical.
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Tabular column:
S.No Current-I
(mA)
Voltage
V2(volts)
Voltage
V3(volts)
Voltage
V1(volts)
V2 +V3…..volts
1
2
Theoretical Calculations:-
Case1. (For measuring Voltage V2 across R1)
REq = (R1+R2)
IEq = V / REq
( V1 ) → V x R1 …………..Volts
(R1 + R2)
Case 2. (For measuring Voltage V3 across R2)
REq = (R1+R2)
IEq = V1 / REq
( V2 ) → V x R2 …………..Volts
(R1 + R2)
Comparison of values Practical with Theoretical
KVL Theorem Voltage
V2(volts)
Voltage
V3(volts)
Voltage
V1(volts)
Theoretical Values
Practical Values
Safety Precautions:
1. Reading must be taken without parallax error.
2. Measuring instruments must be connected properly & should be free from errors.
3. All connections should be free from loose contacts.
4. The direction of currents should be identified correctly
Result:-
Kirchhoff‟s Voltage Law is verified practically.
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3.SERIES AND PARALLEL RESONANCE
AIM: To Obtain Frequency characteristics of the Series and Parallel Resonance circuits and
finding the Resonant Frequency, Bandwidth and Quality factor or RLC Net work.
Theory:- The Resonant circuit is a combination of R,L and C elements, it should be Series, Parallel or
General Sires Parallel circuit.
The Resonance of a circuit is defined as “the condition that occurs when a circuit inductive and
Capacitive Reactances are exactly equal”.
Depending upon the frequency of the source voltage Vs , the circuit behaves either as inductive
or as capacitive . How ever at a particular frequency, when the inductive reactance XL equals
the Capacitive Reactance XC , then the circuit behaves as a purely resistive circuit. This
phenomenon is called Resonance.
These resonant circuits make it possible for a radio receiver to tune in, or receive a desired
frequency to select a particular radio station. The tuning of a knob means operating a variable
capacitance, hence various resonant frequencies can be tuned / obtained by varying the
capacitance in the RLC circuit.
Series Resonance:-
Inductive reactance XL =ωL or 2 πfL
Capacitive Reactance XC = 1/ωC or 1/2 πfC
The total impedance of the circuit = Z = R + J (XL - XC )
But the resultant reactance of the circuit is zero under resonance, XL - XC = 0
At the Resonant Frequency ,the net reactance is zero because XL = XC
The circuit impedance Z becomes minimum and is equal to the Resistance R , and the line
current will be maximum.
Parallel Resonance :-
It is said to be in resonance, when its reactive component must be zero. It is defined as the
frequency where the inductive and capacitive reactances of the two branches are equal and the
line current is a minimum, hence L/RC (dynamic impedance) is maximum and is resistive.
Comparison of Series Resonance & Parallel Resonance
S,No Series Resonance Parallel Resonance
1 It magnifies Voltage It magnifies Current
2 It has unity Power Factor It has unity Power Factor
3 Current at resonance is E/R Current at resonance is E/(L/RC)
4 Effective Impedance is R Effective Impedance is L/RC
5 Resonant frequency is 1/2π√LC Resonant frequency is 1/2π√1/LC-R2/L
2
6 It has minimum Impedance & maximum
current at resonant frequency
It has maximum Impedance & minimum
current at resonant frequency
7 inductive reactance XL ,and the Capacitive
Reactance XC, cancelled mutually
The suspectance of inductance and
capacitance is cancelled mutually
Resonant Frequency (fr) = 1/2π√LC. Hz.
The Quantitative performance of a resonant circuit is evaluated by means of Q-fatcor
Q-Factor = Real Power / Avg. Power = I2 XL / I
2 R = XL / R= ω0 L/R(Induct.React)
Q-Factor = I2 XC / I
2 R = XC / R= 1/ω0 C/R or f0 / (f2-f1)(For capacitive reactance)
Q-Factor of series ckt is defined as “ voltage magnification” of the ckt at Resonance
Volt.Magnification= Voltage across L or C / applied voltage=VL/VR
Band Width = (f2-f1) Hz
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Circuit Diagram:-(using CRO) Circuit Diagram:-(w/o. CRO)
Fig.No.1
Fig.No.2
APPARATUS:
SI.No Equipment Range Qty
1 Function Generator 1
2 CRO 1
3 Inductors 1
4 Capacitors 1
5 Resistors 1
6 Connecting Wires As Reqd
PROCEDURE:for Series Resonance
1. Connect the circuit as shown in the above figure No.1 .
2. By adjusting the Function Generator, Feed a sine wave of 5v (P-P) amplitude
3. Vary the input frequency of FG and observe the output in CRO and also note the output.
peak-to-peak amplitude. or in case of no CRO, simply note down the current flowing
through the circuit which is connected by an Ammeter in the circuit in series.
4. Tabulate these values. and plot the graph between Frequency and current.
5. Draw the graph between input frequencies vs. output voltage.(in case of CRO)
6. From the graph observe the resonant frequency i.e. frequency at witch voltage is maximum.
7. By using graph calculate, cutoff frequency, f1, f2, Band Width and Q-factor.
8. Compare the Theoretical values of F0 and Q-Factor with Practical values. Parallel Resonance
1. Connect the circuit as shown in the above figure .No.2.
2. Repeat the procedure from steps 2 to 7 for parallel resonance.
3. From the graph observe the resonant frequency i.e frequency at witch voltage is maximum.
4. Note down the Voltmeter reading VR , calculate I as , I = VR/R…..mA
5. Calculate the Z of the circuit using Z=Vs/I, where I is obtained in the above step
6. By using graph calculate cutoff frequency,f1,f2, Band Width and Q-factor.
7. Compare the Theoretical values of F0 and Q-Factor with Practical values.
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OBSERVATIONS:
Series Resonance Parallel Resonance
Model Graphs: Series Resonance Parallel Resonance
PRECAUTIONS:
1. Reading must be taken without parallel error.
2. Measureing instruments must be properly calibrated.
RESULT:
The Frequency Response of Series and Parallel Resonance studied and bandwidth,
Quality factor is also calculated.
S.No Frequency
(Hz)
I
(Amp)
Output
voltage
(VR)
Z
1 2 3 4 5 6 7 8 9
10
S.No Frequency
I
(Amp)
Output
voltage(VR)
1 2 3 4 5 6 7 8 9
10
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Ex.No.04
TWO PORT NETWORK PARAMETERS – Z-Y PARAMETERS
AIM:- To study and measure the Z, Y parameters for a given 2 Port Passive Network.
THEORY:- A port is defined as any pair of terminals into which energy is supplied, or from which energy
is withdrawn or where the network variables may be measured.
A 2 port network is a simple network, having inside a rectangular box and the network
has only 2 pairs of accessible terminals usually one pair represents the input, and the another
represents output.
In the above fig having 4 terminals have been paired into ports 1-1’ and 2-2’
Two ports containing no sources in their branch are called passive port.
Two ports containing sources in their branches are called active port.
The voltage and current assigned to each of the two ports.
V1, I1 → input terminals
V2, I2 → output terminals
V1, V2, I1, I2 → are variables (2 of these are dependent variables & 2 independent variables)
The Number of possible combinations generated by the four variables taken 2 at a time, is 6.
Then, there are 6 possible sets of equations describing a 2 port network.
For the 6 combinations, the names of the parameters are chosen to indicate dimensions
(Impedance, admittance) law of consistent dimensions (Hybrid), or the principal application of
the parameter (Transmission).
NAME FUNCTION
EQUATION Express In terms of
Open circuit Impedance V1,V2 I1, I2 V1 = Z11 I1 + Z12 I2
V2 = Z21 I1 + Z22 I2
Short circuit admittance I1, I2 V1,V2 I1 = Y11 V1 + Y12 V2
I2 = Y21 V1 + Y22 V2
Transmission (ABCD) V1,I1 V2,I2 V1 = AV2 - BI2
I1 = CV2 - DI2
Hybrid V1,I2 I1,V2 V1 = H11 I1 + H12 V2
I2 = H21 I1 + H22 V2
Z Parameters (Open circuit Parameters):
Z11, Z12, Z21, Z22 are called Z-parameters of 2 port network.
1) Z11 → Input Impedance where output is open circuited
Z11=V1/I1 where I2=0
2) Z12 → Reverse Transfer Impedance (mutual), when input is open circuited.
Z12 = V1/I2 where I1=0
3) Z21 → Forward Transfer Impedance, when output is open circuited.
Z21 = V2/I1 when I2=0
4) Z22 → Output Impedance, when input is open circuited.
Z22 = V2/I2 when I1=0
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Y Parameters (Short circuit Parameters → Y11, Y12, Y21, Y22)
1) Y11 → Input Admittance where output is short circuited
Y11=I1/V1 where V2=0
2) Y22 → Output Admittance, when input is short circuited.
Y22 = I2/V2 where V1=0
3) Y12 → Reverse Transfer Admittance, when output is short circuited.
Y12 = I1/V2 when V1=0
4) Y21 → Forward Transfer Admittance, when input is short circuited.
Y21 = I2/V2 when V2=0
Transmission or ABCD Parameters:
ABCD are Transmission parameters. These parameters are also known as by other
name, chain parameters. In this system of parameters volt and current at port 1 are
expressed in terms of volt and current at port 2.
1) A = Ratio of input volt to the output voltage when output is open circuited.
A = V1/V2 when I2=0
2) B = Ratio of Input volt to output current when output is short circuited.
B= -V1/I2 when V2=0.
3) C = Ratio of Input current to output voltage when output is open circuited.
C = I1/V2 when I2=0.
4) D = Ratio of Input current to the output current when output is short circuited.
D = -I1/I2 when V2=0.
Z parameters:
Procedure for Z parameters:
1) Connect the circuit as per fig.1
2) Keep the port 2 terminals (C&D)open , then (I2=0).
3) Set desired voltage on V1, between terminal A&B
4) Measure –V2 between terminal C&D and I1 ,Then tabulate V1, V2, I1.
5) Now open the Port1.terminals (A&B ),Connect desired voltage to port 2(C&D)
terminals , (I1=0) as shown in fig.2 then measure V2, V1, I2.
Apparatus required:-
S.No Equipment Range Quantity 1 DC.RPS-Voltage Source 0-30 Volts/2A 1
2 Resistors 1.1KΩ,2.2KΩ,3.3KΩ 3
3 Ammeter-DC 0-200 milliamps 2
4 Voltmeter-DC 0-30/20 volts 1
5 Connecting wires 1.00 sq.mm As required
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Y parameters:
Procedure for Y parameters 1) Connect the circuit as per fig.3, connect desired voltage at port1(A&B). Then short
port2.(C&D) Note the values of I1, I2, V1.
2) Connect any desired voltage at port2.(C&D) and short port1 (A&B) as shown in fig.4
3) Then note the values of V2, I1, I2.
Tabular Column for ‘Z’ parameters: Tabular Column for ‘Y’parameters:
( When I2 = 0 , Port.2 Open ) ( When I/P,Port.1, short circuited, V1 = 0 )
Z11 = V1/ I1 : Z21 = V2/ I1 Y12 = I1/ V2 : Y22 = I2/ V2
( When I1 = 0 , Port.1 Open) ( When O/P,Port.2 ,short circuited, V2 = 0 )
Z22 = V2/ I2 : Z12 = V1/ I2 Y11 = I1/ V1 : Y21 = I2/ V1
V2 V1 I2 Z22 Z12 5V
10V 12V
Procedure for
ABCD(Transmission)Parameters: 1. From the above circuits, take the values of V1 ,V2 ,I1 ,I2 and calculate the A,B,C,D , by
using the above formulas.
2. Compare the Practical values with theoretical values.
Precautions:-
1. Avoid loose connections. 2. Readings should be taken carefully.
3. Get your connected circuit checked by staff member.
Result:-
Z, Y parameters and ABCD Parameters are calculated and verified practically.
V1 V2 I1 Z11 Z21 5V
10V 12V
V2 I1 I2 Y12 Y22 5V 8V
10V
V1 I1 I2 Y11 Y21
5V
8V
10V
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Ex.No.05
SUPER POSITION THEOREM
AIM: To verify the superposition theorem for a given circuit.
Theory:-
This theorem states that “The current through, or voltage across any element in a linear
bilateral network is equal to the algebraic sum of the currents of voltages produced
independently by each sources ( i.e. by getting other sources to 0).
A given response in a network regulating from a number of independent sources(including
initial condition source) may be computed by summing the response to each individual source
with all other sources made in operative( reduced to zero voltage or zero current).
This statement describes the property homogeneity in linear networks.
So it is the combined properties off additivity and homogeneity off linear network.
It is a result of the linear relation between current and volt in circuits having linear impedances.
Apparatus required:-
S.No Equipment Range Quantity
1 DC.RPS-Voltage Source 0-30 Volts/2A 1
2 Resistors 1.1KΩ,2.2KΩ,3.3KΩ 3
3 Ammeter-DC 0-200 m.Amps 1
4 Voltmeter-DC 0-20V or 0-30V 2
5 Connecting wires Single lead As required
Given Circuit
PROCEDURE:-
8. Connect the circuit as per the figure shown above
9. Adjust the V1 voltage as 10 volts, & V2 voltage as 15 volts
10. Measure the current through R3 resistor using Ammeter ie, I total.
11. Now keep the V1 voltage same(ie,10V), & remove V2 voltage and short, then measure
the current through R3 resistor , ie, I’
12. Now keep the V2 voltage as 15 volts, & remove V1 voltage and short, then measure the
current through R3 resistor , ie, I ”
13. Verify that the I total. = I’ + I ”
14. Tabulate the readings in the tabular column
15. Repeat the procedure for deferent voltage values of V1 , & V2
16. Compare the values Practical to Theoretical.
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Theoretical Calculations:-
Case1. (For measuring I‟)When the V1 source is applied (V2 is zero)
REq = (R2 ║ R3)+R1
IEq = V1 / REq
I‟ = IEq X R2
R2+R3
Case 2. (For measuring I”)When the V2 source is applied (V1 is zero)
REq = (R1║ R3)+R2
IEq = V2 / REq
I‟‟ = IEq X R1
R1+R3
Case 3. (For measuring Itotal)When the V1 & V2 source are applied (Original circuit)
Comparison of values Practical with Theoretical
Super Position Theorem Case-1 ( I’ )
m.Amp
Case-2 ( I”)
m. Amp Case-3 ( I-total)
m.Amp
Theoretical Values
Practical Values
Safety Precautions:
5. Reading must be taken without parallax error
6. Measuring instruments must be connected properly & should be free from errors
7. All connections should be free from loose contacts
Result:-
Super position theorem is verified practically.
Theoretical
Calculations.
I(total)=V/R3
(or) = ( I’+I”)
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Ex.No.06
RECIPROCITY THEOREM
AIM: To verify Reciprocity theorem for a given circuit.
Theory:-
Ohm‟s law and Kirchoff‟s laws are the fundamental tools for network analysis, while network
theorems are very powerful tools for solving complicated network problems.
It is applicable only a simple sources network. The theorem states that “In any linear bilateral
network the ratio of voltage source E volts in one branch to the current I in another branch is
the same as the ratio obtained if the positions of E and I are interchanged, other emf‟s being
removed.”. (or)
“If in any network , a potential V introduced in to any branch ‟A‟ causes a current „I‟ to flow in
any other branch „B‟; then the same potential „V‟ introduced into branch „B‟ will cause the
same value of current to flow in branch „A‟.
In other words , this law simply means that “V&I” are mutually interchangeable. The ratio V/I
is called the transfer resistance or Impedance.
Apparatus required:-
S.No Equipment Range Quantity
1 DC.RPS-Voltage Source 0-30 Volts/2A 1
2 Resistors 1.1KΩ,2.2KΩ,3.3KΩ 3
3 Ammeter-DC 0-200 m. Amps 1
4 Voltmeter-DC 0-30 or 20 volts 2
5 Connecting wires single lead As required
Circuit Diagram:-
PROCEDURE:-
1. Connect the circuit as per the figure shown above
2. Apply DC 10 volts in RPS , at AB side (ie,VAB) voltage .
3. Measure the current at XY side using Ammeter or in DMM ie, I XY
4. Now interchange the Supply (ie,VAB) & Ammeter (ie,I XY ), So now VXY =10V
and measure the current at AB side , ie, IAB.
5. Verify that the IAB. = I XY , ie , the both values will be equal
6. Tabulate the readings in the tabular column
7. Repeat the procedure at deferent voltages
8. Compare the Practical values to Theoretical.
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Theoretical Calculations:-
Case1. (For measuring IXY)When the Voltage at AB side applied (ie,VAB)
REq = (R2 ║ R3)+R1
IEq = V1 / REq
I XY = IEq X R3
R2+R3
Case 2. (For measuring IAB)When the Voltage at XY side applied (ie,VXY)
REq = (R1║ R3) +R2
IEq = V2 / REq
IAB = IEq X R3
R1+R3
Compare the Practical values with Theoretical
Reciprocity Theorem Case-1 ( I xy ) m.Amp Case-2 ( IAB) m.Amp
Theoretical Values
Practical Values
Safety Precautions:
1. Reading must be taken without parallax error
2. Measuring instruments must be connected properly & should be free from errors
3. All connections should be free from loose contacts
Result:-
Reciprocity theorem is verified practically.
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Ex.No.07
VERIFICATION OF MAXIMUM POWER TRANSFER THEOREM
Theory:-
Max power will be delivered by network to the load, if the impedance of network is complex
conjugate of load impedance and vice versa
(or)
The maximum transformer states that “ A load will received maximum power from a linear
bilateral network when its load resistance is exactly equal to the Thevenin‟s resistance of
network, measured looking back into the terminals of network. APPARTUS:-
SI. No Equipment Range Qty
1 DC.RPS. Voltage source. 0-30V 1
2 Resistors 4.7KΩ 1
3 Variable Resister 1-10k Ω 1
4 Ammeter-DC 0-20mA 1
5 Voltmeter-DC 0-50V 1
6 Connecting wires Single lead As required
Circuit Diagram:-
PROCEDURE:
1. Connect the circuit as shown in the above figure.
2. Apply the voltage 12V from RPS.
3. Now vary the load resistance (RL) in steps and note down the corresponding Ammeter
Reading ( IL) milli amps and Load Voltage (VL) volts
4. Tabulate the readings and find the power using formula.→ Power = (I2) RS
5 Draw the graph between Power and Load Resistance.
6. After plotting the graph , the Power will be Maximum , when the Load Resistance will be
equal to source Resistance .
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Tabular Column
S.No
IL(mA) VL(Volts) R=VL/IL (Ω) Power(P max)=I2*RL(mW)
1
2
3
4
5
6
7
8
9
10
Theoretical Calculations:-
R = (RS + RL)=470 + 470 = 940 Ω
I = V / R = 12/940 = 0.012765A or 1.27mA
Power = (I2) RS = (0.127)
2 * 470 = 0.07658…..mW
Safety Precautions:
1. Reading must be taken without parallax error
2. Measuring instruments must be connected properly & should be free from errors
3. All connections should be free from loose contacts
Result:-
The Maximum Power Transfer theorem is verified practically.
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Ex.No.08
THEVENIN’S THEOREM Aim:- To verify Thevenin‟s Theorem for given circuit .
Theory:-
Thevenin‟s theorem replaced a complicated circuit with a constant voltage supply and
resistance in series with it.
The Thevenin‟s Theorem states that “Any two terminals linear bilateral DC network can be
replaced by an equivalent circuit consisting of a voltage source Vth in series with all
equivalent resistance Rth”. (OR)
“The current through a Load Resistor “R” connected across any two points A&B of an active
network, containing Resistors and one or more sources of em‟s is obtained by dividing the
Potential Differences between A&B , with R disconnected by (R+r), where „r‟ is the resistance
of the network measured between point A&B , with „R‟ disconnected and source of emf
replaced by their internal Resistances”
Vth → Open circuit voltage between the terminals of network.
Rth →Equivalent resistance measured between terminals.
When all energy sources are replaced by their internal resistances.
APPARTUS:-
SI. No Equipment Range Qty
1 DC.RPS. Voltage source. 0-30V 1
2 Resistors 1.1 KΩ 2
3 Resistors 2.2 KΩ 2
4 Resistors 3.3 KΩ
5 Variable Resister 1k Ω 1
6 Ammeter-DC 0-20mA 1
7 Voltmeter-DC 0-50V 1
8 Connecting wires Single lead As required
Circuit Diagram:-
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PROCEDURE:
1. Connect the circuit as per the circuit diagram.
2. Apply the DC.voltage 10V from RPS.( ie, at AB side)
3. Note down the Load current (IL) from Ammeter.
4. Now remove Load Resistor (RL) & connect a Voltmeter between CD, and measure the
Voltage (ie Vth)
5. Remove the Supply from AB & short.
6. Also remove the voltmeter, measure the Resistance at CD with the help of DMM (ie, Rth)
7. After getting the Vth & Rth ,Now make a circuit as applying Vth voltage and connecting Rth
& keeping Load resistance (RL) as it is in the original circuit and measure load current ((ILth)
through (RL) by connecting a DMM or ammeter in series with (RL).
8. Compare IL, & ILth and observe that the both readings are equal
9. Repeat the procedure at different voltages.
TABULAR COLUMN [Comparing the Theoretical & Practical values]
Thevenin’s
Theorem
Case-1
( IL) m.Amps
Case-2
( Vth) volts
Case-3
( Rth ) KΩ
Case-4
( ILth) m.Amps
Theoretical Values
Practical Values
Conclusion:- Case-1 and Case-4 must be equal
Theoretical Calculations:-
Case-3
Measuring Thevenin‟s Equiv. Resistance → ( Rth )
For given circuit, by removing supply and shorting AB
( Rth ) → (R1 ║R3) + R2
Case-2
Measuring the Thevenin‟s Voltage → ( Vth )
( Vth ) → V x R3 …………..Volts
(R1 + R3)
Case-1
REq = [R3 ║( R2+RL)]+ R1
IT =V / REq I L → IT x R3 ……mA
R3+R2+RL
Case.4
I LTH → Vth / (Rth+ RL)……mA
PRECAUTIONS:
1 Reading must be taken without parallax error.
2. Measuring instruments must be handled properly.
3. All connections should be free from loose contacts
RESULT:
Thevenin‟s Theorem is verified practically
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Ex.No.09
NORTON’S THEOREM Aim:- To verify Norton‟s Theorem for given circuit .
Theory:- It is similar to Thevenin‟s Theorem , while Thevenin‟s Theorem based on the idea of
equivalent source of emf. Norton‟s theorem is based on the idea of an equivalent current source
Norton‟s theorem replaces a complicated circuit with an a constant current supply and
resistance in parallel with it.
This theorem states that “Any two terminals linear bilateral DC network can be replaced by an
equivalent ckt, consisting of a current source Isc in parallel with an equivalent resistance Rth.
(or) Any arrangement of the source of emf „s and the resistance can be replaced by an
equivalent current source in parallel with a resistance .The current from the source is the short
circuit current in the original system , and r is the equivalent resistance of the network between
it‟s two terminals, when all sources of emf‟s are replaced by their internal resistances.
Isc → Short circuit voltage between the terminals of network.
Rth → Equivalent resistance measured between terminals.
When all energy sources are replaced by their internal resistances
APPARTUS:-
SI. No Equipment Range Qty
1 DC.RPS .Voltage source. 0-30V 1
2 Resistors 1.1 KΩ 2
3 Resistors 2.2 KΩ 2
4 Resistors 3.3 KΩ 1
5 Variable Resister 1k Ω 1
6 Ammeter-DC 0-20mA 1
7 Voltmeter-DC 0-30V 1
8 Connecting wires Singe lead As required l
Circuit Diagram:-
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PROCEDURE:
1. Connect the circuit as per the circuit diagram.
2. Apply the DC.voltage 10V from RPS.( ie, at AB side)
3. Note down the Load current (IL) from Ammeter.
4. Now remove Load Resistor (RL) & connect a Ammeter between CD, and measure the
current (ie IN)
5. Remove the Supply from AB & short.
6. Also remove the Ammeter, measure the Resistance at CD with the help of DMM (ie, RN)
7. After getting the IN & RN ,Now make a circuit as applying IN Current source and
connecting RN & keeping Load resistance (RL) as it is in the original circuit and measure
load current ((ILN) through (RL) by connecting a DMM or ammeter in series with (RL).
8. Compare IL, & ILN and observe that the both readings are equal
9. In case of Current source not available, give equivalent DC Supply voltage (ie, IN * RN)
TABULAR COLUMN [Comparing the Theoretical & Practical values]
Norton’s
Theorem
Case-1
( IL) m.Amps
Case-2
( IN) mA
Case-3
( RN )KΩ
Case-4
( ILN) mAmps
Theoretical Values
Practical Values
Conclusion:- Case-1 & Case-4 must be almost equal
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Ex.No.01 Cycle-II
MAGNETIZATION CHARACTERISTICS OF DC SHUNT GENERATOR
(OPEN CIRCUIT CHARACTERISTICS)
AIM: To find the Critical Resistance (RC) and Open Circuit Characteristics (OCC)
of a dc Shunt Generator.
THEORY: MAGNETISATION CURVE:
The graph between the field current and corresponding flux per pole is called magnetization
characteristics of the machine this is same as B-H curve of the material used for the pole construction.
In a dc Generator for any given speed the induced emf in the armature is directly pro portional
to the flux per pole.
Eg = ΦZN / 60 x P / A
Where Φ is the flux per pole in webers, Z is the number of conductors in armature.
OPEN CIRCUIT CHARACTERISTICS: The armature is driven at a constant to its rated value. The terminal voltage (VL) at no – load condition
is measured at diff If values. The graph VL Vs If is called open-circuit characteristics. VL differs from
Eg due to (a) Armature reaction (b) voltage drop in armature circuit Ia is very small at no load
condition, these effects are negligible. Hence VL = Eg at no load condition, thus the O.C characteristics
is same as magnetization curve.
CRITICAL FIELD RESISTANCE (Rc):
Critical field resistance is defined as maximum field circuit resistance at which the Generator would
just excite at any given speed. At this value the Generator will just excite. If the field circuit resistance
is increases beyond this value the generator will fail to excite.
It is the initial slope value of the OCC curve in the linear region (AB) passing through the origin. If
the field circuit resistance (Rf) is increase to Rc, the machine fail to excite and no emf is induced in the
generator for exciting the Generator (Rf < Rc ).
CRITICAL SPEED: For any given field circuit resistance, the speed above which the Generator builds up an appreciable
voltage is called Critical speed.
CIRCUIT DIAGRAM:
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NAME PLATE DETAILS:
MOTOR GENERATOR
Power 3.7KW Power 3.7KW
Speed 1500RPM Speed 1500RPM
Type of Wound Shunt Type of Wound Shunt
Armature voltage 220V Armature voltage 220V
Armature current 20. A Armature current 17 A
Field voltage 220 V Field voltage 220 V
Field current 0.8 A Field current 0.8 A
APPARATUS REQUIRED:
S.No Name Type Range Qty
1 Voltmeter MC 0-300 V 01
2 Ammeter MC 0-2 A 02
3 Tachometer Digital 0-10000RPM 01
4 Rheostat WW 360Ω/1.7A 01
5 Connecting wires 2.0 2 mm As req
PROCEDURE:
Initially note down the name plate details of DC shunt Generator
1. Note down the Name plate details of DC shunt motor & Generator
2. Connect the circuit as shown in circuit diagram.
3. Keep the Field rheostat of DC motor in minimum resistance position
4. Keep the Potential divider in minimum voltage position ( ie, maximum resistance)
5. Switch ON the MCB and close the DPST for dc voltage 220 V to the circuit.
6. Start the motor with the help of 3 point starter, by slowly varying (cutting) the starter
resistance.
7. By adjusting the Field rheostat of the Motor, to get the rated speed (1500rpm).
8. By varying Potential Divider gradually note down the OC voltage (Eo) and field current up
to 150% of rated voltage of generator & tabulated.(increasing level)
9. By reducing, Potential Divider gradually note down the OC voltage (Eo) and field current
up to minimum position of generator & tabulated. (decreasing level)
10. Keep the field rheostat of the motor in original position,ie, minimum position,
11. Open the DPST switch and Next Push OFF the MCB, to disconnecting supply Mains.
TABULAR COLUMN:
S No If (Amps) Increased(Rise) ↑
(Eg)Volts
Decreased (Fall)↓
(Eg)Volts
1 0.05 2 0.10 3 0.15 4 0.20 5 0.25 6 0.30 7 0.35 8 0.40 9 0.45
10 0.50 11 0.55 12 0.60 13 0.65
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MODEL GRAPH:- Field current Vs Terminal Voltage ( If Vs Eg )
( for finding the Critical Field Resistance)
For finding the Field Resistance (circuit diagram)
PROCEDURE:-
1. Connect the circuit as shown in the below fig.
2. Keep the Armature winding open.
3. Vary the input DC Supply from the controller and note the ammeter & voltmeter readings.
4. V/I Ratio will give the field resistance.
5. Repeat the same for deferent input voltages, and find out RF
6. The average value gives the Field Resistance RF.
Precautions:
1. Before switch ON , keep the potential divider at maximum position , & motor field
rheostat at minimum position.
2. Before switch OFF , follow the replica procedure.
3. Through out the experiment, the generator should be runned at no-load condition.
4. Always avoid the loose connection in the circuit.
RESULT:
Determined the Critical Field Resistance (Rc) and drawn the Graph:
. Field current Vs Terminal Voltage (ie, If Vs Eo.)
Sl.No Vf
Volts
If
Amps
Rf=Vf/If
Ohms
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Ex.No.2
SWINBURNE’S TEST
AIM: To per-determine the efficiency of a DC shunt machine
(Considering it as a motor or generator) by performing Swinburne‟s test and also to draw the
characteristics curve.
THEORY: Testing of DC machines can be divided in to three methods; (1) Direct method (2)
Regenerative and (3) Indirect method.
Swinburn‟s Test is an indirect testing method of DC machine. In this method, the
constant losses of the DC machine is calculated at no load. Hence it is efficiency either as a
motor are as generator can be pre determined. In this method, the power requirement is very
small. Hence, this method can be used to predetermine the efficiency of higher capacity dc
machines as motor and as generator. It is applicable to those machines flux is practically
constant, ie, Shunt and compound machines only
Power input at no load = Constant losses + Armature copper losses
Power input at no load = constant losses
Power input = Va Ia + Vf If
LOSSES IN DC MACHINE:
The losses in a dc machine can be divided as (1) Constant losses (2) Variable losses, which
changes with the load.
CONSTANT LOSSES:
1. Mechanical losses: Friction and winding losses are called mechanical losses. They
depend up on the speed. A dc shunt machine is basically a constant speed machine
booth as generator or as motor thus, the mechanical losses are constant.
2. Iron losses: For a DC shunt machine, the field current Hence the flux per pole is
constant (Neglecting the armature reaction which reduces the net flux in the air gap).
Hence, hysterics and eddy current losses (Which are also called as iron losses) remain
constant.
3. Field copper losses: Under normal operating condition of a dc shunt machine, the
field current remains constant. Thus, power received by the field circuit (Which is
consumed as field copper losses) is constant.
Constant losses in a dc shunt machine = Mechanical losses + Iron losses + Field copper losses
Variable losses: The power lost in the armature circuit of a dc machine increases with the
increasing load. Thus armature copper losses are called as variable losses.
Efficiency of a dc machine: Output power
% Efficiency = --------------------X 100
Input power
As a Generator Input Power, Pin = Pout + constant losses + armature copper losses at a given
load Ia2 Ra.
Pout = VL IL →watts
Where Ia = IL + If → Self excited generator
Ia = IL → Seperately excited generator
As a motor input power Pin = VL*IL + Vf If
Output power Pout = Pin –Constant losses – armature copper losses
IL = Ia + If → Self excited motor
IL = Ia → Seperately exciter motor
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NAME PLATE DETAILS:
S.No MOTOR RANGE
1 Power 5.0 H.P
2 Armature voltage 230V
3 Armature current 19 A
4 Field current 0.8 A
5 Field voltage 220 V
6 Speed 1500 RPM
7 Type of wound Shunt
APPARATUS:
S.No. Name Type Range Qty
1 Voltmeter MC 0-300V 01
2 Ammeter MC 0-10/20A 01
3 Ammeter MC 0-2A 01
4 Tachometer Digital 0-10000rpm 01
5 Rheostat W.W 360Ω/1.7A 01
6 Connecting Wires pvc 1.0 2mm As reqd
CIRCUIT DIAGRAM:
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PROCEDURE:
Initially note down the name plate details of DC shunt machine
1. Connect the circuit as shown in the Circuit diagram.
2. Keep the field rheostat of DC Shunt motor at minimum position.
3. Switch ON the MCB & Close the DPST for 220V DC supply to the circuit.
4. Start the motor with the help of 3 point starter, by slowly varying (cutting) the starter
resistance
5. By varying (adjusting) the field rheostat ,Run the motor at Rated speed .
6. Don‟t apply the load to the motor (no load condition).
7. Now , note down the readings of all meters.
8. Keep the field rheostat in original position, after that open the DPST .
9. Finally Switch OFF the MCB.
TABULAR COLUMN:To find out the constant losses
S.N
o
Ter
min
al
Volt
age
(V)
No l
oad
curr
ent
I L(a
mps)
Fie
ld c
urr
ent
If(a
mps)
No l
oad
Arm
.curr
ent
I ao=
I L-I
f(am
ps)
Spee
d (
N)
RP
M
Const
ant
loss
es
VI 0
-Ia 0
2R
a
Input
=V
L*I L
(W)
Outp
ut
2π
NT
/60(W
)
Eff
icie
ncy
( %
η )
OP
/IP
*100
1
2
3
4
5
6
7
8
9
10
CALCULATION:
1. Motor:-
No Load IL = IA + IF ; IA = IL-IF :
Fraction of Load = X
Load current(IL) = X x IN
Input (Wi) = VL* IL (Watts)
Arm.Cu. Losses(Wcu) = Ia2
* Ra
Constant Losses (Wc)= VI0-Ia02Ra
(WT)Total Loss = Arm.Cu.Loss + Const.Loss
Out put = Input-Total Loss(Wi-WT)
% Efficiency (η) = Out Put /Input *100
2. Generator:- No Load IA = IL+IF :
Fraction of Load = X
Load current(IL) = X x IN
Out put (Wo) = VL* IL (Watts)
Arm.Cu. Losses(Wcu) = Ia2 * Ra
Constant Losses = (Wc)= VI0-Ia02Ra
Input(Wi) = Out put + Total Losses(Wi+WT)
Total Losses (WT)= Arm. Cu. Losses +
Const.Losses
% Efficiency (η) = Out Put /Input *100
S.No V IA IF
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MODEL GRAPH:
1. Load current IL Vs %Efficiency (Motor / Generator)
Armature Resistance Test:
Measure the field resistance and armature resistance by conducting another simple
circuit. as given below
PROCEDURE:
1. Connect the circuit as shown in the fig,
2. Motor shaft should not rotate.
3. Vary the input voltage from 0-50 V
4. Note down ammeter and voltmeter readings and enter in the tabular column.
5. Calculate the Resistance = V /I
Tabular Column:
Sl.No Arm. Volt (VA) Arm.
Current(IA)
Arm. Resistance(RA) = (VA) /
(IA)
1
2
3
4
For measuring the Field Resistance, simply replace A- AA with F - FF, remaining all are
same.
Precautions:
1. Do not make loose connections.
2. Before giving supply field Rheostat should be in minimum Resistance position.
3. Through out the experiment, the motor should be run at “no load condition”.
RESULT:
Efficiency of DC machine, ie, Generator/ Motor are obtained.
Efficiency of Generator is higher than the motor.
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Ex.No.3 BRAKE TEST ON A DC SHUNT MOTOR
AIM: To perform brake test on the given dc shunt motor and obtain the performance
characteristics of the motor.
THEORY:
In this method the motor is subjected to direct mechanical loading by attaching a
brake drum and water cooled pulley to the motor shaft. A rope or belt is wound the pulley at its
two ends. The two end are connected to two spring balances S1 and S2 . The tension of the belt
can be adjusted by tightening it on the pulley. The tangential force on the pulley is equal to the
difference of the two spring balance readings.
Tangential Force = (S1 – S2) Kg.m.
Torque on the pulley = 9.81 (S1 – S2) * R . N-m. (where g=9.81cm)
Where R is the radius of the pulley in meters and S1 & S2 are spring balance readings in Kg.
Power output = [2πNT / 60] Watts.
Where N is the R.P.M.
Power input = VI Watts where V → is the motor input voltage, I→ the motor input current.
Efficiency = (Output / Input) X 100
NAME PLATE DETAILS:
Type of Wound Shunt
Armature Voltage 230V
Armature Current 20A
Field Current 220V
Speed 1500RPM
Power 5H.P
APPARATUS:
S.No Apparatus Type Range Qty
1 Ammeter MC 0-20A 1
2 Voltmeter MC 0-300V 1
3 Rheostat WW 360Ω/1.7A 1
4 Tachometer Digital 0-10000 1
5 Connecting Wires PVC 2.0 Sq.mm As reqd
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CIRCUIT DIAGRAM:
PROCEDURE:
1. Connect the circuit diagram as shown in fig.
2. Keep the field regulator Rsh at minimum resistance value.
3. Check that the belt on the pulley is free so that there is no load on the pulley.
4. Start the motor slowly using the starter.
5. Adjust the field current so that the motor runs at its rated speed.
6. Apply load on the pulley gradually in steps, tightening the belt around it.
7. Take the readings of the ammeter and voltmeter connected to the motor and input the
two spring balance readings and the speed at every step.
8. Cool the pulley throughout the loading period by pouring water.
9. Continue the experiment till full load of the motor is reached.
10. Tabulate the observation.
TABULAR COLUMN:
Radius of pulley R = _________mts
S.No VL
Volts
IL
Amps
N RPM
S1
Kg
S2
Kg
T
Torque
I/P
watts
O/P
watts
%η
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CALCULATIONS:
Radius of pulley Drum = 100mm or 0.1 mts
Torque T = 9.81 (S1 – S2) x R N-m
Power Output = 2πNT/60 Watts
Power Input = VIL Watts
% efficiency = Output / Input * 100
MODEL GRAPH:
PRECAUTIONS:
1. Cool the pulley while the experiment is performed.
2. While measuring the radius of the pulley effective radius must be considered.
RESULTS: Draw the performance characteristics for the DC shunt motor on a graph sheet.
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Ex.No.4
OPEN CIRCUIT & SHORT CIRCUIT TEST ON 1-Φ TRANSFORMER.
(OC & SC Test)
AIM:
To determine the Iron Losses, Copper Losses of a Transformer and hence calculate its
Regulation & Efficiency.
THEORY:
A transformer is a static device which transfers the electrical energy from one circuit to another
circuit without any change in the frequency. The transformer works on the principle of
electromagnetic induction between two windings placed on a common magnetic circuit. These
two windings are electrically insulated from each other and also from the core.
The losses in a transformer are (1) magnetic losses or core losses
(2) Ohmic losses or copper losses
The losses of a transformer, magnetic losses and ohmic losses can be determined by
performing (a) open circuit test and (b) short circuit test. From the above tests, the efficiency
and regulation of a given transformer can be predetermined. The power consumed during these
tests is very less as compared to the load test. In this experiment LV side parameters are
denoted by suffix 1 and HV side parameters by suffix 2.
Open circuit test:
In open circuit test, usually HV side is kept open and meters are connected on LV side,
the ammeter reads the no load current Io and wattmeter reads the power input Wo. The no load
current Io is 2 to 5 %of full load current .hence the copper losses at no load are negligible. Wo
represents the iron losses or core losses. Iron losses are the sum of hysterisis and eddy current
losses.
Wo=Vo Io CosФo Cos Фo=Wo/(Vo Io)
IW=Io Cos Фo Iµ=Io Sin Фo
Ro=Vo/Iw Xo=Vo/Iµ
Short circuit test:
This test is performed to determine the equivalent resistance and leakage reactance of the
transformer and copper losses at full load condition.
In this test usually LV side is shorted and meters are connected on HV side .a variable low
voltage is applied to the HV winding with the help of an auto transformer. This voltage is
varied till the rated current flows in the HV side or LV side .The voltage applied is 5 to 10% of
rated voltage, while the rated current flows in the windings. The watt meter indicates the full
load copper losses at Vsc. But the iron losses at this low voltage are negligible as compared to
the iron losses at rated voltage.
Wsc=full load copper losses =I2
HV Req (hv) =I22
R02
Z02 =Vsc/Isc X02 =√ (Z 022-R02
2)
TRANSFORMER RATINGS:
Power: 2 KVA
Primary/secondary 110V /220V
Isc = Rating of Transformer / Applied voltage = 2000/220 = 9.09Amps
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APPARATUS:
S.No Apparatus Type Range Qty
1. Ammeter M.I 0-2 A 01
2. Voltmeter M.I 0-300V 02
3. LPF Wattmeter Dynamometer 0-200 Watts 01
4. UPF Wattmeter Dynamometer 0-3 KW 01
5. 1-ΦVariac Fully variable 0-270V/15A 01
6. Ammeter M.I 0-20A 02
CIRCUIT DIAGRAM: Fig (a) Open circuit tests.
Fig (b) Short circuit test
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PROCEDURE:
O C Test:
1. Connect the circuit diagram as shown in the figure.
2. Gradually increase the Auto Transformer voltage till the voltmeter reads 230V.
3. Record the voltmeter, ammeter and UPF watt meter readings.
4. The ammeter indicates the no load current and wattmeter indicates the iron losses.
5. Set the auto transformer at zero position and switch off the supply.
S C Test:
1. Connect the circuit diagram as shown in the figure.
2. Gradually increase the auto transformer voltage till the ammeter reads rated current of
the transformer on HV side
3. Record the voltmeter, ammeter and UPF Wattmeter readings.
4. The ammeter indicates Isc (short circuit current), voltmeter indicates Vsc (short circuit
voltage), and wattmeter indicates Wsc, copper losses of the transformer at full load
condition.
5. Set the auto transformer in zero position and switch off the supply.
OBSERVATIONS:
O.C test: Voc= Ioc= Wo= (LV Data)
S.C test: Vsc= Isc= Wsc= (HV Data)
CALCULATIONS:
Calculation of Ro and Xo equivalent circuit from OC and SC test
Vo= Io= Wo=
Iron losses =Wo= VoIoCosФo CosФo=Wo/ (VoIo)
Iw=IoCosФo Ro=Vo/Iw.
Iµ=Io SinФo Xo=Vo/Iµ.
HV side Ro= LV side RoxK2
HV side Xo= LV side XoxK2 where K=VHV/VLV
Calculation of R01 and X01 for equivalent circuit:
Isc= Vsc= Wsc=
Full load copper losses or variable losses=Wsc=Isc2 R02
R02=Wsc/Isc2 Z02=Vsc/Isc
X02=√ (Z022-R02
2)
R01=R02/K2 X01=X02/K
2
Calculation of % Regulation from SC test:
% Regulation=I2 (R02CosФ+X02SinФ)/V2
Efficiency calculation:
Output in KVAxP.fx100
%η= Output in KVA x P.f + Iron losses +X2 Full load copper losses
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Model graphs:
Efficiency Vs load
%Regulation Vs P.f
PRECAUTIONS:
1. Before switching on the supply check if the auto transformer reads zero.
2. In SC test don‟t allow the current to go beyond the rated value.
3. Before switching off the supply set the auto transformer to zero position.
4. In OC test apply only the rated voltage initially.
CONCLUSION:
Iron losses, copper losses and hence % regulation and efficiency of a single phase
transformer are determined by performing OC and SC test.
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Exp.No.5
LOAD TEST ON SINGLE PHASE TRANSFORMER
AIM:To conduct load test on 1-Φ transformer and to obtain % efficiency & Regulation. Apparatus Required:
Sl.No Apparatus Type Range Quantity 1 Auto Transformer 230V/0-270 V 1
2 1-ΦTransformer 1 KVA - 220/220V 1
3 Voltmeter MI (0-300)V 2
4 Ammeter MI (0-10)A 2
5 Wattmeter UPF 0-300/5A 2
6 Resistive Load 230 V / 10A 1
7 Connecting wires pvc 1.00 sq.mm As reqd
Name Plate Details Single Phase Transformer Primary Voltage : 220 V : Secondary Voltage : 220 V Capacity : 1 KVA or 2KVA : Frequency : 50 Hz
THEORY:
When the secondary is loaded the secondary current I2 is setup. The magnitude and phase of I2 with respect to V2 is determined by the characteristics of the load. The secondary current sets up its own mmf and hence its own flux ф2 which is in opposition to main primary flux ф which is due to I0 the secondary ampere turns N2*I2 are known as demagnetizing ampere turns .The opposing secondary flux I2 weakens the primary flux Φ momentary. Hence primary back Emf E1 tends to be reduced. For a movement V1 gain the upper hand over E1 and hence causes more current to flow in primary.
Let the additional primary current be I21 .It is known as load component of primary current. This current is antiphase with I21 the additional primary mmf N1*I2 sets up its own flux Φ21 which is in opposite to Φ2 and is equal to its magnitude. Hence the two cancel each other out. So the magnetic effects of secondary current I2 are immediately neutralized by the additional primary current I21.Hence whatever the load conditions be, the net flux passing through core is approximately the same as no-load.
Circuit Diagram: LOAD TEST ON 1-Φ TRANSFORMER
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PROCEDURE: 1. Connect as per the circuit diagram 2. Close the DPST switch 3. Adjust the Auto transformer till the rated voltage is reached ie,220V 5. Note down the readings of primary voltmeter, ammeter and wattmeter & secondary
voltmeter, ammeter and wattmeter 6. Apply load in steps and note down the corresponding readings till the rated current
is reached.ie, 1000V/220V = 4.545 Amps (say up to 4 Amps) PRECAUTIONS: 1. The autotransformer should be kept at minimum voltage position, before starting. 2. Before switching off the supply the 1-Φ Auto Transformer (variac) should be
brought back to “0” ie, minimum voltage position. Tabular Column:-
Sl.No
Primary
Volt. V1
Primary Current
I1
Input
Power W1 x mf
Sec. Volt V2
Sec.
Current I2
Output Power
W2 x mf
%
Reg.%
= E0 – V X 100 V
Model Calculations: % Efficiency η = Output power X 100
Input power
% Regulation = E0 – V X 100 V
Graphs: 1.Output power Vs efficiency 2. Output power Vs % regulation
RESULT:
Thus the load test on single phase transformer was performed and the respective graphs were plotted.
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