Ee Practical in PDF

Vidharbha Youth Welfare Society’s

Prof Ram Meghe College of Engineering and Management

New Express Highway, Badnera – Amravati

First Year Engineering Department

CERTIFICATE

This is certify that Mr./Miss______________________________________________________________

Class Roll No.________________ Section__________________ has attended the practical classes and

completed the practical work satisfactorily in ________________________________________________

for __________________semester and __________________group as prescribed by Sant Gadge baba

Amravati University, Amravati during the year ____________________ with grade__________________

Head of Department Prof. Incharge

Date:

ELECTRICAL ENGINEERING PRACTICAL GROUP B

PROF. RAM MEGHE COLLEGE OF ENGINEERING AND MANAGEMENGT Page 2

Vidharbha Youth Welfare Society’s

Prof Ram Meghe College of Engineering and Management

New Express Highway, Badnera – Amravati

INDEX

Sr.No. Name of Experiment Date Remarks

1 ___________________________________________________

2 ___________________________________________________

3 ___________________________________________________

4 ___________________________________________________

5 ___________________________________________________

6 ___________________________________________________

7 ___________________________________________________

8 ___________________________________________________

9 ___________________________________________________

10 ___________________________________________________

11 ___________________________________________________

12 ___________________________________________________

13 ___________________________________________________

14 ___________________________________________________



EXPERIMENT NO. 1

AIM: Verification of Kirchhoff’s laws.

SIGNIFICANCE: These laws are applicable in the analysis of almost all the electrical and electronic circuits.

APPRATUS: 1) Connecting wires, 2) 4mm patch cords, 3) Kirchhoff’s laws kit, 4) Digital Multimeter – 3 nos. 5) Regulated DC power supplies

THEORY:

THE KIRCHHOFF’S VOLTAGE LAW It states that “the algebric Sum of all the voltages aroud closed loop is zero”. IR DROP:- Whenever we move in the direction of current there is a drop in voltage, since the current always flows from point at higher potential to the point at lower potential. Hence Voltage drop in the current direction is taken as negative and vice- versa. SIGN CONVENTIONS: If we move from negative terminal of source to positive terminal, there is a voltage rise. Therefore it is to be considered positive. If we move from +ve terminal to –ve terminal of voltage source, there is voltage drop. Therefore it is to be considered as –ve. B) THE KIRCHHOFF’S CURRENT LAW (KCL) It states that “At any instant the algebraic sum of all the currents entering or leaving a node is zero”. A node is a common point in a circuit where more than one elements are connected. If we are going to assign negative sign to the incoming current then we will have to assign positive sign to the outgoing current and vice-versa.



CIRCUIT DIAGRAM:

Kirchhoff’s Voltage Law:

Kirchhoff’s Current Law:



SPACE FOR CIRCUIT DIAGRAM (IF THERE IS CHANGE IN DIAGRAM GIVEN ABOVE):

PROCEDURE:

A) For Kirchhoff’s Voltage law:

1) Connect the power supply cord of the kit to the socket. 2) Connect variable dc source in the circuit 1 of Kirchhoff’s voltage law. 3) Switch on the supply and measure voltage across all the three resistances i.e R1, R2, and R3 with

the help of multimeter of figure 1. Also find out the polarity across each resistor i.e whether there is a voltage rise or voltage drop and mark the positive and negative sign sign across each resistor. This you can done with the help of multimeter, if the meter displays reading without any sign that means the point at which red wire has been connected is positive and point at which black wire has been connected is negative. And if the meter displays reading with negative sign before it that means, the point at which red wire has been connected is at negative and the point at which black wire has been connected is at positive. So, now mark the positive and negative signs across each resistors in the circuit diagram.

4) Note down the readings of voltages across all the resistors in the observation table. 5) Note down 3 more readings in the same manner as described above by varying the supply

voltage level. 6) Verify kirchhoff’s voltage law for one or two readings, by writing down KVL equation and verify it

in the space provided for calculations.



B) For Kirchhoff’s Current law:

1) Connect the power supply cord of the kit to the socket. 2) Connect variable dc source in the circuit 2 of Kirchhoff’s current law. 3) Switch on the supply and measure current through all the three resistances i.e R1, R2, and R3 or

through all the three branches with the help of multimeter(for this set multimeter on 10A range). Also find out the direction of current through each resistors. This you can done with the help of multimeter, if the meter displays reading without any sign that means that the current is flowing from red to black wire or in the direction of red to black. And if the meter displays reading with negative sign before it that means, the current is flowing from black to red wire or in the direction of black to red. Mark the direction of current in the figure 2 by arrow in that branch in the direction of current

4) Note down the readings of current through all the resistors in the observation table. 5) Note down 3 more readings in the same manner as described above by varying the supply

voltage level. 6) Verify kirchhoff’s current law for one or two readings, by writing down KCL equation and verify it

in the space provided for calculations.

OBSERVATIONS:

For KVL:

S.N Voltage across source, Vs (V)

Voltage across R1, V1 (V)



1

2

3

4



For KCL:

S.N Current through R1, I1

(A) Current through R2,

I2 (A) Current through

R3, I3 (A)

1

2

3

4

SPACE FOR OBSERVATIONS (IF THERE IS CHANGE IN OBSERVATIONS GIVEN ABOVE):



CALCULATIONS:

RESULTS/CONCLUSION:

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

PRECAUTIONS:

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________



EXPERIMENT NO. 2

AIM: Verification of Superposition Theorem.

SIGNIFICANCE: Network theorems are used to speed up the analysis of circuits. This theorem is very useful in the analysis of electric circuits containing more than one source.

APPRATUS: 1) 2mm Patch cords, 2) Digital Multimeter, 3) Superposition Theorem Kit.

THEORY:

Statement: Superposition theorem states that "In any linear bilateral network containing two or more sources, the response in any element is equal to the algebraic sum of the responses caused by individual sources acting alone, while the other sources are non-operative i.e., while considering the effect of individual sources, other ideal voltage sources and ideal current sources in the network are replaced by short circuit and open circuit across their terminals”.

EXPLANATION: 1. Select only one source and replace all other sources by there internal resistances. (If the source is the ideal current source replace it by open ckt. if the source is the ideal voltage source replaces it by short ckt.) 2. Find the current and its direction through the desired branch. 3. Add all the branch currents according to their direction to obtain the actual branch current.



CIRCUIT DIAGRAM:

FOR BRANCH c-d:



FOR BRANCH e-f:




PROCEDURE:

For branch c-d:

1) Connect the power supply cord of the kit in the socket. 2) Connect 12 V source in branch a-b i.e +12V at upper point and ground at the lower point in

branch a-b with the help of 2mm patch cords. 3) Connect 5 V source in branch g-h i.e +5V at upper point and ground at the lower point in branch

g-h with the help of 2mm patch cords. 4) Now short the open path in branch e-f with the help of patch cord. 5) Connect milliammeter in branch c-d. 6) Switch on the supply and note down the value and direction of the current in branch c-d. Mark

this direction in the diagram. This current will be Icd 7) Now replace 12 V source with a short with patch cord and keep only 5 V source active. Again

measure the value and note down direction of current in the branch c-d. this current will be Icd’. If the direction of current changes mark the direction same as before but note down reading with negative sign (draw another diagram for this condition).

8) Now connect 12 V source in the circuit replace 5 V source with a short. Again measure the value and note down direction of current in the branch c-d. this current will be Icd’’. If the direction of current changes mark the direction same as before but note down reading with negative sign (draw another diagram for this condition).

9) Take the algebraic summation of Icd’ and Icd’’ and compare it with Icd, both the values should come same. Do the calculation and verify superposition theorem.



For branch e-f:

Repeat the procedure as written above for branch e-f. for this connect milliammeter in the branch e-f and short the branch c-d.

OBSERVATIONS:

Condition Current flowing through

branch c-d Current flowing through

branch e-f

When both the sources are acting

Icd = _______ Ief =_______

When only 5V source is acting

Icd’ =_______ Ief’ =_______

When only 12V source is acting

Icd’’ =_______ Ief’’ =_______




CALCULATIONS:

RESULTS/CONCLUSION:

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

PRECAUTIONS:

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________



EXPERIMENT NO. 3

AIM: Verification of Thevenin’s Theorem.

SIGNIFICANCE: Network theorems are used to speed up the analysis of circuits. This is very important theorem for analyzing circuits.

APPRATUS: 1) 2mm Patch cords, 2) Digital Multimeters, 3) Superposition Theorem Kit.

THEORY:

STATEMENT: It states that any linear active two terminal network containing resistance and voltage sources and /or current sources can be replaced by single voltage source VTH in series with a single resistance RTH. The thevenin equivalent voltage VTH is the open circuit voltage at the network terminals, and thevenin resistance RTH is the resistance between the network terminals when all the sources are replaced by their internal resistances. EXPLANATION: a) Steps to find the RTH • Open circuit current sources and short circuit voltage sources. • Open circuit the load resistance • Find out the equivalent resistance which is RTH across the open circuit terminals. b) Steps to find the VTH • Open circuit the load resistance • Find out the open circuit voltage VTH across the open circuit terminals by using mesh analysis. c) Find out the load current IL by connecting the load resistance to the given circuit. d) Draw the equivalent circuit. Connect the voltage source VTH in series with RTH and RL .



CIRCUIT DIAGRAM:

THEVENIN’S EQUIVALENT DIAGRAM:




PROCEDURE:

1) Set the value of RL at one fixed value in both original and thevenin’s equivalent network same and equal. 2) Connect +12V, DC power supplies at their indicated position using patch cords. 3) Measure voltage between terminals 2 & 4 using voltmeter, for this connect terminal 2 to the + terminal of DC Voltmeter and 4 to –ve terminal. It is the required practical value of Thevenin's equivalent voltage (VTH). 4) Disconnect the patch cord between terminals 1 to +12V and Gnd to Gnd. 5) Connect test point 1 & Gnd (of circuit) so as to replace source by its internal resistance (assuming it negligible). 6) Measure resistance between terminals 2 & 4 using multimeter. It is the required practical value of Thevenin's equivalent resistance RTH. 7) Connect a 2mm patch cord between terminals 1 and supply and Gnd to Gnd socket. 8) Connect an ammeter between terminals 2 and 3 to measure load current IL flowing through load resistance of original circuit. 9) Now connect 2 mm patch cords between +5v supply and terminal 5 of thevenin’s equivalent network and ground to ground. 10) Connect milliammeter between terminals 6 and 7 of equivalent network and measure the load current IL flowing through load resistance RL of thevenin’s equivalent network. 11) Compare load current (IL) flowing through both of the load resistances and also with the theoretical values of IL calculated.

OBSERVATIONS:

1) Value of RL set in original and thevenin’s equivalent network = ________

2) Practical measured value of VTH = __________

3) Practically measured value of RTH = __________

4) Practical value of IL measured from original network = ___________

5) Practical value of IL measured from thevenin’s equivalent network = ___________

6) Theoretically calculated value of RTH = ________

7) Theoretical calculated value of VTH = ________

8) RL set (this same value should be used in theoretical calculations) = _______

9) Theoretically calculated value of IL = ________




CALCULATIONS:



RESULTS/CONCLUSION:

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

PRECAUTIONS:

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________



EXPERIMENT NO. 4

AIM: Verification of Maximum Power Transfer Theorem.

SIGNIFICANCE: Network theorems are used to speed up the analysis of circuits. This is very important theorem for analyzing amplifier electronic circuits.

APPRATUS: 1) 2mm Patch cords, 2) Digital Multimeters, 3) Superposition Theorem Kit. THEORY:

DC Circuit:

The maximum power transfer theorem states that “maximum power is delivered from a source resistance to a load resistance when the load resistance is equal to the thevenin’s resistance as seen from load terminals.” RTH = RL is the condition required for maximum power transfer.

AC Circuit:

a. The maximum power transfer theorem states that maximum power is delivered from a source impedance to load impedance when the load impedance is equal to the complex conjugate of the thevenin’s impedance as seen from load terminals.

b. The maximum power transfer theorem states that maximum power is delivered from a source impedance to load resistance when the load resistance is equal to the magnitude of the thevenin’s resistance as seen from load terminals.”



CIRCUIT DIAGRAM:


PROCEDURE:

1) Connect +12V, DC power supplies at their indicated position using patch cords. 2) Measure voltage between terminals 2 & 4 using voltmeter, for this connect terminal 2 to

the + terminal of DC Voltmeter and 4 to –ve terminal. It is the required value of Thevenin's equivalent voltage (VTH).

3) Disconnect the patch cord between terminals 1 to +12V and Gnd to Gnd.



4) Connect test point 1 & Gnd (of circuit) so as to replace source by its internal resistance (assuming it negligible).

5) Measure resistance between terminals 2 & 4 using multimeter. It is the required value of Thevenin's equivalent resistance RTH.

6) Switch off the kit and set the value of RL less than RTH by connecting ohmmeter (multimeter) between terminals 2 and 3.

7) Now connect ammeter (multimeter) between terminals 2 and 3, voltmeter between terminals 3 and 4 and measure value of IL and VL. Also calculate RL and power for this reading.

8) Now increase RL such that it should become more than what we calculated in step 7 but less than RTH, this is going to be second reading. Once again read values of VL and IL from multimeter and calculate RL and power.

9) Now set RL = RTH with the help of multimeter. And measure value of VL and IL and also calculate power for this reading.

10) Increase RL now it will become greater than RTH, note down reading of VL and IL and calculate RL and power for this reading.

11) Further Increase RL , note down reading of VL and IL and calculate RL and power for this reading.

12) Find out value of RL at which maximum power is getting transferred to load.

OBSERVATIONS:

S.N Voltage across

load VL (V) Current through

load IL (mA)

Load resistance

RL (kΩ) Power PL (mW)

1

RL < RTH

2

3 RL = RTH

4

RL > RTH

5




CALCULATIONS:



GRAPHS: Draw graph between load resistance RL on X-axis and power on Y-axis



RESULTS/CONCLUSION:

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

PRECAUTIONS:

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________



EXPERIMENT NO. 5

AIM: Verification of line and phase relationship of current and voltage in a balanced Star connected load.

SIGNIFICANCE:

APPRATUS: 1) Connecting wires, 2) A.C Voltmeter (0-600V) – 3no. 3) A.C Voltmeter (0-300V) – 3no. 4) Ammeter MI (0-5A) – 3no. 5) Three phase autotransformer – 1no. 6) Rheostat (5A, 100Ω) – 3nos.

THEORY:

The three phases of the three phase system can be used independent of each other supplying

power to different loads. Each load gets power from a single phase. This system is adopted for domestic

power supply.

The loads are connected to form a three phase system. The loads (i.e., impedances Z1, Z2 and

Z3) can then be either star-connected or delta-connected. Such a load is called a three phase load. In a three

phase load, if all the three impedances are equal (both in resistive as well as reactive parts), the load is said to

be a balanced load. An example of such a balanced three-phase load is a three-phase motor, which has three

identical windings. To such a load, if a balanced three phase supply is applied, the currents will also be

balanced. Conversely, if it is carrying balanced currents, the voltages across the circuit will also be balanced.

In a three phase system, there are two sets of voltages: (i) line voltages, and (ii)

phase voltages. Similarly, there two sets of currents: (i) line currents, and (ii) phase currents.

Star Connected System:

In figure 1, VRN is the r.m.s value of the voltage drop from R to N. That is, this is the

phase voltage of phase R. Thus, VRN, VYN and VBN denote the set of three phase voltages. The term, line

voltage is used to denote the voltage between two lines. Thus VRY represents line voltage between the lines

R and Y. The other line voltages are VYB and VBR. To determine the relation between phase voltages and line

voltages, we analyse the phasor diagram, shown in figure 2





CIRCUIT DIAGRAM:




PROCEDURE:

1) Connect the circuit as shown in figure. 2) Connect the meters of proper ratings. 3) Initially keep the autotransformer at zero position. Then gradually increase the voltage and look at

reading of current in all the three ammeters. If current is different change the load in each phase accordingly and bring current in all the three meters equal. This is the balanced condition means now load is balanced. Now don’t change the position of load throughout the experiment.

4) Note down the readings of all the meters in observation table. 5) Now change the voltage level by changing autotransformer position and again note down reading of

all the meters. 6) Take 5 to 6 readings like this. 7) Do the calculations and verify relationship between line and phase voltages and current in star

connected load. 8) Draw the graph for at least two readings.

OBSERVATIONS:

S.N Voltmeter Reading Ammeter Reading Ratio between line and phase

voltages

VRN

(V) VYN

(V) VBN

(V) VRY

(V) VYB

(V) VYB

(V) IR

(A) IY

(A) IB

(A) VRY/ VRN VYB/ VBN VYB/ VRY

1

2

3

4

5

6




CALCULATIONS:



GRAPHS:



ii)



iii)



RESULTS/CONCLUSION:

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

PRECAUTIONS:

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________



EXPERIMENT NO. 6

AIM: Verification of line and phase relationship of current and voltage in a balanced Delta connected load.

SIGNIFICANCE:

APPRATUS: 1) Connecting wires, 2) A.C Voltmeter (0-600V) – 3no. 3) Ammeter MI (0-10V) – 3no. 4) Ammeter MI (0-5A) – 3no. 5) Three phase autotransformer – 1no. 6) Rheostat (5A, 100Ω) – 3nos.

THEORY:

The three phases of the three phase system can be used independent of each other supplying

power to different loads. Each load gets power from a single phase. This system is adopted for domestic

power supply.

The loads are connected to form a three phase system. The loads (i.e., impedances Z1, Z2 and

Z3) can then be either star-connected or delta-connected. Such a load is called a three phase load. In a three

phase load, if all the three impedances are equal (both in resistive as well as reactive parts), the load is said to

be a balanced load. An example of such a balanced three-phase load is a three-phase motor, which has three

identical windings. To such a load, if a balanced three phase supply is applied, the currents will also be

balanced. Conversely, if it is carrying balanced currents, the voltages across the circuit will also be balanced.

In a three phase system, there are two sets of voltages: (i) line voltages, and (ii)

phase voltages. Similarly, there two sets of currents: (i) line currents, and (ii) phase currents.

Delta connected System:

Let IR’R, IB’B, IY’Y, be the r.m.s values of the phase currents in the three windings of

the generator. Their assumed posistive directions are indicated by arrows in figure 1. Since the load is

assumed balances, these currents are equal in magnitude and differ in phase by 120˚, as shown in the phasor

diagram of figure 2, therefore we can write

|IR’R| = |IY’Y| = |IB’B| = Iph

From figure 1, it can be seen that the phase current IR’R flows towards the line conductor R, whereas the

phase current IB’B flows away from it. Applying KCL at the terminal R, we can write

IR = IR’R - IB’B

Above vector addition of IR’R and -IB’B is shown in the phasor diagram. From the symmetrical geometry of the

diagram, it is evident that the linne currents are equal in magnitude and differ in phase by 120˚, Also

IR = 2(Iphcos30˚) = √3Iph



Hence, for a delta connected system with balanced load, the magnitude of line currentand of phase current

are related as

IL = √3Iph

From figure, it is obvious that the line voltage VRY is same as the phase voltage VR’R. hence, for a delta

connected system, we have

VL = Vph



CIRCUIT DIAGRAM:




PROCEDURE:

1) Connect the circuit as shown in figure. 2) Connect the meters of proper ratings. 3) Initially keep the autotransformer at zero position. Then gradually increase the voltage and look at

reading of current in all the three ammeters connected in three phases of delta connected load. If current is different change the load in each phase accordingly and bring current in all the three meters equal. This is the balanced condition means now load is balanced. Now don’t change the position of load throughout the experiment.

4) Note down the readings of all the meters in observation table. 5) Now change the voltage level by changing autotransformer position and again note down reading of

all the meters. 6) Take 5 to 6 readings like this. 7) Do the calculations and verify relationship between line and phase voltages and current in delta

connected load. 8) Draw the graph for at least two readings.

OBSERVATIONS:

S.N Voltmeter Reading Ammeter Reading Ratio between line and phase

voltages

IR

(V) IY

(V) IB

(V) IRY

(V) IYB

(V) IYB

(V) VRY VYB VBR IRY / IR IYB / IY IYB / IB

1

2

3

4

5

6




CALCULATIONS:



GRAPHS:

i)



ii)



iii)



RESULTS/CONCLUSION:

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

PRECAUTIONS:

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________



EXPERIMENT NO. 7

AIM: Study of series R-L-C circuit.

SIGNIFICANCE:

APPRATUS: 1) Connecting wires, 2) A.C Voltmeters (0-300V) – 1 no. 3) A.C Ammeter (0-5A) – 1 no. 4) Wattmeter (10A, 300V) – 1 no. 5) A.C Voltmeters (0-300V) – 3 no.

THEORY:

An a.c voltage of r.m.s value ‘V’ when applied to a R-L-C series circuit establishes an r.m.s current

‘I’ given by

I = V/Z, where Z = √R² + (XL - XC)²

The circuit is as shown in figure 1 and the phasor diagram in figure 2, for the case when XL > XC. The effect of

XL > XC is a lagging power factor, since the current I lags behind voltage V by an angle Ф,

Where Ф = tan^-1 ((XL –XC/R))

If XC > XL then the power factor of the circuit is leading and current I leads the voltage V by an angle Ф.

In an inductor, the copper losses take place due to resistance of its coil, and the core losses take place in the

magnetic core (in case of ‘ironed cored’ inductors).

In the capacitor, the losses take place in the dielectric medium used for making it.



CIRCUIT DIAGRAM:




PROCEDURE:

1) Connect the circuit as shown in figure. Connect resistor, inductor and capacitor in series. Keep the resistor, inductor and capacitor at one position.

2) Initially keep the autotransformer at zero position. Now gradually increase the voltage and bring it upto one level.

3) Note down the readings of current, voltages across resistor, inductor and capacitor, applied voltage and also note down power.

4) Now change the value of resistance or inductance or capacitance or all in the circuit and again note down the readings of all the meters.

5) Do the calculations as given in the calculation table and write down the result. 6) Draw the graph for at least three set of readings. 7) Calculate power factor of the circuit from the graph also.



OBSERVATIONS:

S.N Applied

Voltage V (V)

Current I (A)

Voltage across

resistor VR (V)

Voltage across

inductor VL (V)

Voltage across

capacitor VC (V)

Total Power P (W)

1

2

3

4

5

6




CALCULATIONS:

S.N V = √VR² + (VL – VC)² S =VI p.f, cosФ = P/VI R = P/I² Z = V/I

1

2

3

4

5

6

S.N R = VR/I XL = VL/I XC = VC/I Z = √R² + (XL - XC)² p.f, cosФ = R/Z p.f from phasor

diagram

1

2

3

4

5

6



GRAPHS:

i)



ii)



iii)



iv)



RESULTS/CONCLUSION:

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

PRECAUTIONS:

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________



EXPERIMENT NO. 8

AIM: Study of Parallel R-L-C circuit.

SIGNIFICANCE:

APPRATUS: 1) Connecting wires, 2) A.C Voltmeters (0-300V) – 1 no, 3) A.C Ammeter (0-2A) – 3 no, 4) Wattmeter (10A, 300V) – 1 no. 5) A.C Ammeter (0-5A) – 1 no. THEORY:

Figure 1 shows the parallel RLC circuit. Writing KCL equation, we get total current I supplied by

the applied voltage as

I = IR + IL + IC =

+ V/jXL + V/-jXC = VG + V(-jYL) + (jYC)

Or I = V[G + j(Yc - YL)] = VY

Where, y is the complex admittance of the given circuit. Obviously,

G =

; Y = 1/XL = 1/ωL and YC = 1/XC = ωC/1 = ωC

Note that +j is associated with YC (and not with XL) and –j with YL. The complexadmittance of the circuit can

be written as

Y = G + j(Yc - YL) = √ G² + (Yc - YL)² ∠tan^-1



CIRCUIT DIAGRAM:




PROCEDURE:

1) Connect the circuit as shown in figure. Connect resistor, inductor and capacitor in parallel. Keep the resistor, inductor and capacitor at one position.

2) Initially keep the autotransformer at zero position. Now gradually increase the voltage and bring it upto one level.

3) Note down the readings of applied voltage, total current and current through resistor, inductor and capacitor and also note down power.

4) Now change the value of resistance or inductance or capacitance or all in the circuit and again note down the readings of all the meters.

5) Do the calculations as given in the calculation table and write down the result. 6) Draw the graph for at least three set of readings. 7) Calculate power factor of the circuit from the graph also.

OBSERVATIONS:

S.N Applied

Voltage V (V)

Total Current I

(A)

Current through

resistor IR (V)

Current through

inductor IL (V)

Current through

capacitor IC (V)

Total Power P (W)

1

2

3

4

5

6




CALCULATIONS:

S.N R =

VR/IR XL = VL/IL

XC = VC/IC

Y = √G² + (YC - YL)² Y = I/V p.f from phasor

diagram

1

2

3

4

5

6



S.N I = √IR² + (IL – IC)² S =VI p.f, cosФ = G/Y R = P/I² p.f, cosФ = P/VI



GRAPHS:

i)



ii)



iii)



RESULTS/CONCLUSION:

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

PRECAUTIONS:

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________



EXPERIMENT NO. 9

AIM: Study of Resonance in R-L-C series circuit.

SIGNIFICANCE:

APPRATUS: 1) Connecting wires, 2) Function Generator – 1no. 3) A. C Voltmeter (0 – 150V) – 1no. 4) Digital Multimeters. 5) A.C Ammeter (0 – 5A) – 1no. 6) A. C Voltmeter (0 – 50V) – 3no.

THEORY:

A RLC series network is said to be in resonance when the applied voltage and current are in phase and the frequency at which this phenomena occurs is known as resonance frequency fr. Resonance occurs when inductive reactance is equal to capacitive reactance i.e. XL = XC.

Resonant frequency fr =

∏√

At resonance the impedance is pure resistance .At resonance frequency the current in the circuit is maximum which is given by Imax. BAND WIDTH: A band of frequencies at which the current is 1

√2 times its maximum value. At that

instant the power delivered to the circuit is half of the power at resonance. Hence they are called as half power frequencies f1 and f2. The frequency f1 is termed as lower cut off frequency and f2 is termed as upper cut off frequency. And hence the difference between the two half power frequencies is known as Band-width. B.W = f2 - f1

QUALITY FACTOR: The ratio between resonance frequency fr to bandwidth B.W. It is also given as the ratio of capacitor or inductor voltage at resonance to supply voltage.

Q factor =

or

Q factor =



CIRCUIT DIAGRAM:


PROCEDURE:

1) Connect the circuit as shown in figure. 2) Set the voltage of the signal from function generator to 5V.

3) Vary the frequency of the signal from 100 Hz to 1KHz in steps and note down the corresponding ammeter readings.

4) Observe that the current first increases & then decreases in case of series resonant circuit & the value of frequency corresponding to maximum current is equal to resonant frequency.



5) Note down this frequency and also value of current at resonant frequency. Also note down voltage across inductor, capacitor and resistor at every instant.

6) Draw a graph between frequency and current & calculate the values of bandwidth & quality factor.

OBSERVATIONS:

S. No. Frequency

(Hz)

Current

(mA)

Voltage

across

resistor VR (V)

Voltage across

inductor VL (V)

Voltage across

capacitor VC (V)

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.




CALCULATIONS:

1) Resonant frequency fr =

∏√

2) B.W = f2 - f1

3) Q factor =

or

4) Q factor =





GRAPHS:

i)



ii)



iii)



RESULTS/CONCLUSION:

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PRECAUTIONS:

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EXPERIMENT NO. 10

AIM: Measurement of power and power factor in a single phase ac circuit.

SIGNIFICANCE: i) Active power can be measured for any type of load. ii) Power factor and Reactive power can be calculated if we also measure voltage across and current flowing through the load.

APPRATUS: 1) Connecting wires, 2) Ammeter (0-5A) – 1nos. 3) Voltmeter (0-300V) – 1nos. 4) Wattmeter (10A, 300V) – 1 nos. 5) Load bank. 6) Decade resistance, capacitance, inductance box – 1 each. 7) Autotransformer.

THEORY:

The active power or average power or real power in any single phase load can be measured with the help of dynamometer type wattmeter. The wattmeter contains two coils current coil and pressure or voltage coil. Current coil (CC) is connected in series with the load and pressure coil (PC) is connected in parallel with the load. Wattmeter has four terminals M, L, C, V. The terminal M is connected to mains supply, terminal L is connected on the load side, terminal V is connected to the other end of the load and terminals M and C are short circuited. If we also connect the ammeter and voltmeter along with the wattmeter, we can also measure the voltage across the load and current through it and the product of voltage and current will give apparent power (VI). After knowing active and apparent power we can easily calculate power factor of the load.



CIRCUIT DIAGRAM:




PROCEDURE:

1) Connect the circuit as shown in the circuit diagram using connecting wires and the measuring instruments.

2) Keep the dimmerstat at minimum position and gradually increase the value of voltage at one level by looking at voltmeter.

3) Initially keep the switch of load bank open, at this time current and power will be zero. Take the reading of all the three meters.

4) Now close the switch of load bank, and increase the load gradually by closing one switch at a time.

5) Every time when you increase the load, adjust the voltage to its initial value using autotransformer, if it gets changed.

6) Note down the reading of all the meters, every time when you increase the load by closing one or two switch at a time.

7) Calculate power and power factor using above given formula.

OBSERVATIONS:

S.N Voltage

(V) Current

(A) Active

Power,P (W) Apparent

Power S = VI Power Factor,

cosФ

1

2

3

4

5

6




CALCULATIONS:

Active Power, P = VIcosФ,

Apparent Power, S = VI,

Power Factor, cosФ = (Active Power/Apparent Power) = P/S.



RESULTS/CONCLUSION:

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PRECAUTIONS:

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EXPERIMENT NO. 11

AIM: To study starting of dc shunt motor using three point starter and reversing of dc shunt motor.

SIGNIFICANCE: i) Need of starter at the time of starting. ii) Reversing of DC shunt motor.

APPRATUS: 1) Connecting wires, 2) DC shunt motor, 3) Three point starter, 4) Variable DC power supply (30A)

THEORY:

DC shunt motor is widely used motor in industrial applications. It is a constant speed motor idealy but there some drop in speed from no-load to full-load practically. The dc shunt motor is started with the help of three point starter. Starter consists of resistances in series with armature which limits the starting current of motor and the resistance is gradually cut-out as the motor gains speed. Once the motor is attaining full speed all resistances of starter has been cut out and motor run at rated speed. Since the back emf of motor is zero at the time of staring as motor is at rest so the heavy current can flow through armature and to limit this starter is used. The direction of dc shunt motor can be reversed or changed either by interchanging its field terminals or armature terminals but not both, because changing both the field and armature terminals will not change the direction of motor and the motor will continue to run in same direction.



CIRCUIT DIAGRAM:






PROCEDURE:

1) Connect the circuit as shown in figure 1. Positive of supply is connected to field point F of motor through starter and negative connected to point FF of field.

2) Now switch on the dc supply source and move the starter arm gradually. The motor will pick up the speed slowly and finally runs at rated speed.

3) Note the direction of motor. 4) Now switch off the supply and interchange the field winding terminals F-FF as shown in figure 2,

negative of supply connected to F of field and positive now connected to FF of field through starter. 5) Again move the starter arm gradually. The motor will pick up the speed slowly and finally runs at

rated speed. 6) And again note down the direction of motor. 7) Now switch off the supply and interchange the armature winding terminals A-AA as shown in figure

3, negative of supply connected to A of armature and positive now connected to AA of armature through starter.

8) Again move the starter arm gradually. The motor will pick up the speed slowly and finally runs at rated speed.

9) And again note down the direction of motor.

10) Disconnect the circuit after finishing experiment.

OBSERVATIONS:

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RESULTS/CONCLUSION:

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PRECAUTIONS:

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EXPERIMENT NO. 12

AIM: Determination of voltage and current ratio of single phase transformer.

SIGNIFICANCE:

APPRATUS: 1) Connecting wires, 2) A.C Ammeter (0-5A) – 2nos. 3) Voltmeter (0-300V) – 2nos. 4) Load bank. 5) Autotransformer.

CIRCUIT DIAGRAM:




PROCEDURE:

1) Connect the circuit as shown in the circuit diagram using connecting wires and the measuring instruments.

2) Gradually increase the voltage applied to primary using autotransformer and set it to one level by looking at voltmeter.

3) Initially keep the switch of load bank open, at this time current will be zero. Take the reading of all the three meters.

4) Now close the switch of load bank, and increase the load gradually by closing one switch at a time. 5) Every time when you increase the load, ajdust the primary side voltage to its initial value using

autotransformer. 6) Note down the reading of all the meters, every time when you increase the load. 7) Calculate the voltage and current ratios.



OBSERVATIONS:

SR NO.

PRIMARY SIDE SECONDARY SIDE Voltage Ratio, V1/V2

Current Ratio, I2/I1

Voltage V1 (V)

Current I1 (A)

Voltage V2 (V)

Current I2 (A)

1

2

3

4

5




CALCULATIONS:

VOLTAGE RATIO = V1/V2,

CURRENT RATIO = I1/I2,

RELATIONSHIP:

=

RESULTS/CONCLUSION:

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PRECAUTIONS:

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EXPERIMENT NO. 13

AIM: Determination of Efficiency and Regulation of single phase Transformer by direct loading.

SIGNIFICANCE: From this test we can find out at which load transformer has maximum efficiency and by how much amount secondary side voltage changes on loading a transformer.

APPRATUS: 1) Connecting wires, 2) A.C Ammeter (0-5A) – 1nos. 3) A. C Ammeter (0-10A) – 1nos. 4) A.C Voltmeter (0-300V) – 2nos. 5) Load bank. 6) Autotransformer. 7) Wattmeter (10A, 300V) – 2nos.

THEORY: In a practical transformer there are two types of losses: (1) Cu loss (2) Core/Iron loss. Therefore output of a transformer is always less than input of the transformer. Here transformer is loaded with a variable resistive load. Input to the transformer can be found out by using a wattmeter and output can also be measured by a wattmeter or with the help of voltmeter and ammeter. Input power to transformer = Reading of wattmeter W1 Output power from transformer = Reading of wattmeter W2 % efficiency η = (Output Power / Input Power) × 100% = (W2 / W1) × 100% Voltage regulation(V.R) is the change in the magnitude of secondary voltage from no load to desired load. This change is expressed as a percentage of the full load voltage.

% V.R = ( )

( ) X 100 %



CIRCUIT DIAGRAM:




PROCEDURE:

1) Connect the circuit as shown in figure. 2) Initially keep the dimmerstat or autotransformer at zero position. And also keep the switch of the

load bank open. Now, gradually increase the voltage and bring it upto rated voltage. Note down this reading of voltage on both the side of transformer. This reading of voltage on secondary side is open circuit secondary voltage.

3) Now close the switch of the load bank and gradually increase the load. Every time when load is increased, note down the readings of voltage, current and power on both the sides of the transformer in observation table.

4) Calculate percent voltage regulation and percent efficiency by formula as given in calaculations.

OBSERVATIONS:

Open circuit secondary voltage V20 = ___________

S.N

Primary Side Secondary Side

Voltage V1 (V)

Current I1 (A)

Power P1 (wattmeter

reading x multiplying

factor) (W)

Voltage V2 (V)

Current I2 (A)

Power P2

(wattmeter reading x

multiplying factor) (W)

1

2

3

4

5

6




CALCULATIONS:

S.N % Voltage Regulation =

X 100 % Efficiency =

X 100

1

2

3

4

5

6



GRAPHS: i) Plot graph between output power P2 on X-axis and % Voltage Regulation on Y-axis.



ii) Plot graph between output power P2 on X-axis and % efficiency on Y-axis.



RESULTS/CONCLUSION:

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PRECAUTIONS:

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