EXPERIMENT NO: - 01
DATE:-22-01-2021
Name : Megha Mundra
Roll No. : 19BEE056
AIM: - To perform the procedure for extension of range for ammeter and voltmeter.
APPARATUS:1. Ammeter 0-3 A
2. Ammeter 0- 10 A
3. Lamp bank
THEORY: The range of ammeters and voltmeter is limited on account of certain factors. One of the factors
(except in M-I type) is the current carrying capacity of the control springs. Therefore it
is necessary for measurement of high current and voltage, to employ a device which reduces
the current and voltage to the range of the instrument by known proportion.
There are four types of devices for extending the range of the measuring instruments: (1) Shunts
(2) Multipliers
(3) Current transformers and
(4) Potential transformers.
Shunts and current transformers are used for extending the range of ammeters while multipliers
and potential transformers are used for voltmeters. Shunts and multipliers can be used for both
D.C. and A.C. measurements whereas current transformers and potential transformers can only
be used for A.C. measurements.
• EXTENSION OF RANGE FOR AN AMMETER.
For increasing the range of an ammeter, a suitable value of precise, low resistance known as
shunt is connected in parallel with the ammeter as shown in figure-1. This shunt provides a
bypath for extra current because it is connected across the instrument. Such
shunted instruments can be made to record currents many times greater than their normal fullscale deflection current. The ratio of maximum current (with shunt) to the full-scale deflection
current (without shunt) is known as the ‘multiplying power’ or ‘multiplying factor of the shunt.
Let Rm = resistance of the ammeter
Rsh = resistance of shunt
Im = full-scale deflection current
I = current to be measured
As the ammeter coil and shunt are connected in parallel,
Im * Rm = Ish *.Rsh
= (I - Im) * Rsh
= IRsh-ImRsh
 I * Rsh = Im * (Rm + Rsh)
(I /Im) * Rsh = Rm + Rsh
Rsh * N = Rm + Rsh
Rsh * (N-1) = Rm
Rsh = Rm / (N-1) =Rm/(I / Im-1)
Where N is known as multiplying factor of the shunt.
•
EXTENSION OF RANGE FOR A VOLTMETER:Let V = voltage to be measured
Rm= resistance of the voltmeter
v = voltage drop across the instrument for full-scale deflection.
= Im * Rm
Rse = series resistance
For increasing the range of the voltmeter, suitable value of high resistance is connected in series
with the voltmeter as shown in figure.
It is seen from the diagram, that
Potential difference across Rse = V - v
Im * Rse = V - v
Dividing both sides by v
(Im * Rse) / v = (V / v) -1
(Im * Rse) / (Im * Rm) =N-1
Rse = Rm * (N-1)
Obviously, larger the value of Rse, greater the voltage multiplication.
CIRCUIT DIAGRAM
PROCEDURE:For Ammeter:
1. Calculate the value of shunt resistance for the required extension of the range of the
ammeter.
2. Connect the shunt resistance in parallel with the ammeter as shown in the circuit diagram.
3. Set the different values of the current and verify the range.
4. Similarly repeat the above procedure for another shunt.
For Voltmeter:
1. Calculate the value of the multiplier for the required extension of the range of the
voltmeter.
2. Connect the multiplier in series with the voltmeter as shown in the circuit diagram.
3. Apply the different voltages and verify the range.
4. Similarly repeat the above procedure for another multiplier.
OBSERVATIONS :For Ammeter:- Rm =ammeter coil resistance = 1ohm
SR.
AMMETER
ACTUAL
MULTIPLYING
NO
READING
READING
FACTOR
Im (Amps)
I (Amps)
N = I/Im
1
0.015
0.02
1.33
2
0.03
0.04
1.33
3
0.06
0.07
1.167
4
0.095
0.12
1.263
For Voltmeter:- Rm =Voltmeter coil Resistance = 114.7 kohm
SR.
VOLTMETER
ACTUAL
MULTIPLYING
NO
READING
VOLTAGE
FACTOR
v (volts)
V (volts)
N = V/v
1
0.8
8
10
2
2.1
15.5
7.381
3
3.1
24
7.742
4
4.7
32.5
6.915
CALCULATION:For Ammeter:Im = ___0.015 A_________
I = ___0.02A________
N = I/Im = __1.33_________
Rsh = Rm / (N - 1)
= ____3.03 ohm________
For Voltmeter:v
= ____0.8 V________
V
= _____8V_______
N = V / v = ___10_________
Rse = Rm * (N - 1)
= ___1032.3 Kohm_________
CONCLUSION:In this experiment we learned to extend the range of ammeter and voltmeter using shunt and
multiplier respectively . The range of an ammeter is extended by connecting a low resistance
, called shunt , connected in parallel with ammeter and the range of voltmeter is extended by
connecting a high resistance called multiplier in series with the voltmeter and calculated the
value of Rsh and Rse and observed the deflection in the theoretical and calculated values.
QUIZ:1. An ammeter shunt is merely a __low value of__ resistance that is placed in parallel with
the coil circuit of the instrument in order to measure fairly _large__ currents.
2. The shunts are made of material such as _copper___, which has __low___ temperature
coefficient of resistance.
3. What are the conditions those are to be fulfilled by multiplier?
Ans : The conditions that a multiplier must fulfill are :The resistance of the shunt resistor must not change with time, if the resistance of the shunt is
to change over time the accuracy of the multiplier shall also deplete over time.
The resistance of the shunt must also only vary by a small magnitude with the change in
temperature, again if the resistance changes with the temperature the readings of the
multiplier shall not be accurate or uniform over the temperature gradient
Ideal material to be used are constantat and manganin.
4. Draw the diagram of multi range ammeter.
EXPERIMENT NO: - 02
Name – Megha mundra
DATE: - 09-02-2021
roll – 19bee056
AIM: - To analyze three phase reactive power using one wattmeter method.
APPARATUS:1.
A.C. Supply, 3-Phase, 415V, 50Hz, 15 A
2.
3-Phase variac 415V, 15A
1
3.
Wattmeter 0-1500 W, 0-600 V, 0-5 / 10A
1
4.
Loading Inductive Coil 415 V, 10A
1
5.
A.C. Ammeter 0-5 A
1
6.
A.C. Voltmeter 0-600 V
1
THEORY: In case of balanced three phase circuits, it is simple to use a single wattmeter to read the reactive
power. The current coil of the wattmeter is connected in one line and the pressure coil is
connected across the other two lines as shown in circuit diagram.
Current through the current coil = IR= IL= IPH
Voltage across the pressure coil = VBY = VL = √ 3 VPH
Therefore, reading of wattmeter W= VBY *IR * cos (90 + Φ)
W = √ 3 VPH * IPH * cos (90+Φ)
W= -√ 3 VPH *IPH * sin Φ
Now,
Total reactive volt amperes of the circuit, Q = 3 VPH*IPH * sin Φ
= (- 3) * reading of wattmeter
= (- 3) * W
–1
-1
Phase angle, Φ = tan (Q/P) = sin [W/ √ 3 VPHIPH] = sin -1 [W/ VL IL]
PROCEDURE: 1. Make connections as per circuit diagram.
2. Keep operating dial of 3-phase variac at its null (‘0’) position & then switch on the 3phase a.c. supply.
3. By operating the dial of 3-phase variac, set the input line voltage at 80 Volts.
4. Adjust the rotating arm/lever of 3-phase choke in anticlockwise sense to reduce the choke
coil phase current to some arbitrary value, which is below than it’s rated current. i.e. 10
Amps.
5. Note readings of Wattmeter, Voltmeter & Ammeter.
6. By operating the dial of 3-phase variac, increase the input line-to-line voltage from 80
Volts onwards in steps of 40 volts upto 200 Volts.
7. For each operating line to line voltage note down the readings of Wattmeter, Ammeter &
Voltmeter.
8. Repeat the procedure from step 3 to step7, for different phase currents of 3-phase choke
coil.
9. Operate the rotating arm/lever of 3-phase choke coil in anticlockwise sense to increase
load current otherwise operate it in clockwise sense to reduce load current.
10. A Care should be taken while performing the experiment that the load current flowing
through each phase of choke coil should never exceed for long time above the normal
/rated current i.e. 10 Amps.
11. After noting down all readings, set the operating dial of 3-phase variac at its null (‘0’)
position & then switch off 3-phase a.c. main supply.
OBSERVATIONS :-
Reactive
Power
Q = √3 W
(VAR)
Sin φ
= W
√3 VL IL
Power
factor
angle
φ
(degre
e)
280
484.97
0.624
38.61
2.40
440
762.10
0.588
36.01
215
2.87
620
1073.87
0.580
35.45
250
2.37
850
1472.24
0.582
35.62
Sr.
No.
Line voltage
VL = √3VPH =VYB
(Volts)
Line current
IL = IPH = IR
(Amp.)
Wattmeter
Reading
W
(Watts)
1.
140
1.85
2.
180
3.
4.
CALCULATIONS:3-ф Reactive Power (theoretically) Q = √3 VL IL sin ф = 439.96 VAR
3-ф Reactive Power (practically)
3-ф Active Power
Q = √3 W = 440 VAR
P = √3 VL IL cos ф = 747.497 W
3-ф Apparent Power S = 3 VL IL = 1296 W
RESULT:-
3-Phase reactive power consumed by 3-phase star connected balanced load
calculated
is
______439.96__VAR______
(theoretically)
and
_______440_VAR________(practically).
CONCLUSION:- In conclusion to this experiment, we can say that there only a minute
difference between the theoretical and practical value. We also observe that the φ
After the conclusion of this experiment we got to know about the various components of
power present in the AC circuits, the conditions for their existence and the factors
affecting them. We also verified our knowledge by practically performing the experiment
and measuring the power first theoretically and then verifying practically using one
wattmeter
QUIZ: 1. What do you mean by active and reactive power?
Sol – Active and reactive power are terms used to define energy flowing in the electrical
circuit. Active power is the power that is actually consumed by the load (measured in Watts),
it is the power that the consumers pay for. For a purely resistive circuit the current and voltage
are in phase and thus the power delivered in active. Whereas, in a purely reactive load, the
current and the voltage are out of phase, thus the power flows to and from between the source
and the load. It is basically the power that flows back and forth in the circuit without doing
any useful work (measured in Voltage-ampere reactive or VAR).
2. Draw the phasor diagram for the 3-phase reactive power measurement by one wattmeter
method?
3. What are the others method for measurement of Reactive Power?
Sol - Other methods for the measurement of reactive power are –
1. Single phase Two wattmeter method
2. Polyphase wattmeter method
3. By using a power meter
4. Using a power quality meter
5. Three phase two wattmeter method
6. Straightaway displaying the reactive power using meter.
LEARNING OUTCOME - the take away from this experiment was to understand the concept
of reactive power, how to calculate it using one wattmeter methods, its derivation and
calculation
EXPERIMENT NO: 03
Name : Megha Mundra
DATE: 3-02-2021
Roll No. : 19BEE056
AIM: - To measure medium resistances using Wheatstone bridge.
APPARATUS:1. Wheatstone bridge.
2. Unknown resistance
3. Galvanometer
4. Connecting wires
THEORY:The classification of resistance from measurement point of view is as follows :
1. Low resistance: All resistances of the order of 1 ohm as under.
2. Medium resistance: Resistances from 1 ohm to about 0.1 mega ohm.
3. High resistance: Resistance of the order of 0.1 mega ohm and higher.
Wheatstone bridge is a very important device used in the measurement of medium resistance. It is
still accurate and reliable instrument and is extensively used in industries. The Wheatstone bridge
is an instrument for making comparison measurements and operates upon null indication principle.
This means the indication is independent of the calibration of the null indicating instruments or any
of its characteristics. The figure shows the schematic diagram of Wheatstone bridge. These basic
circuits has four resistive arms consisting or resistances R1, R2, R3 and R4 together with a source of
EMF and null detector usually a galvanometer G or other sensitive current meter. The bridge is said
to be balanced when there is no current through the galvanometers or when the potential differences
across the galvanometer is zero. This occurs when
R3*R1 = R2*R4
R1= (R2 * R4)/R3
Where R1 = Unknown resistance
R3 = Standard arm resistance.
R2 = Standard arm resistance
R4 = Standard arm resistance
Note: - Accuracy of 0.1% is quite common with a wheatstone bridge as opposed to accuracy of 3%
to 5% with ordinary method.
PROCEDURE:➢ Connect the circuit diagram as shown in figure.
➢ Set suitable ratio of (R2/R1).
➢ Obtain balance condition by adjustment of standard resistance R3.
➢ Note down the value of ratio of R2/R1 and R3.
➢ Calculate the value of X for different values of resistances.
CIRCUIT DIAGRAM :
OBSERVATIONS:Sr
No.
Ratio
R3
Measured Resistance
(R2/R1)
(Ω)
X = R3 * (R2/R1)
Actual
resistance
(Ω)
(Ω)
% Error.
1.
5
15
75
72
4.16
2.
3
20
60
57
5.26
3.
6
12
78
69
13.04
4.
4
10
40
36
11.11
CONCLUSION : Wheatstone bridge method is used to determine the resistance of the unknown resistor, the
resistances of the other three are adjusted and balanced until the current passing through the
multimeter decreases to zero. We observed that there is difference in the actual and the measured
resistance value due to aging of instruments, human errors like parallax and personal errors that
includes poor techniques and carelessness and other instrumental errors.
QUIZ :1. In Wheatstone bridge method of measuring resistances, a resistor of unknown resistance
is balanced against _standard variable_ of known _resistance__.
2. Why the Wheatstone bridge method is more accurate than ohmmeter method.
A Wheatstone bridge is more accurate than ohmmeter method of measuring resistance because
it uses null method. The main advantage is that at null point current does not flow in arm of the
galvanometer, so internal resistance of the cell and resistance of galvanometer does not affect
the null point. It is a convenient and easy method in comparison to other methods. We can also
calculate the unknown resistance by Ohm’s law in which we need to calculate the least counts and
reading of ammeter and voltmeter. Here we have to measure the currents and potential
differences across all components of the circuit which makes the method difficult and time taking.
3. What are the limitations Whetstone Bridge?
The limitations of Wheatstone Bridge are as follows:
•
This method is not precise when it is used to measure large resistance. When large
resistance is measured the results are non-linear.
•
The value of resistance of bridge degrades due to due to the heating effect of the current
through the resistance.
•
The galvanometer is insensitive to imbalance when the value of resistance to be
measured is too high.
•
For low resistance measurement, the resistance of the leads and contacts becomes
significant and introduces an error.
LEARNING OUTCOME:
While performing this experiment we studied about electrical bridges and also derived a
mathematical equation for electrical bridges and then we learnt about the Wheatstone bridge
method for medium resistance measurement which is one of the most accurate and precise method
in comparison to other methods. Although this method also has few disadvantages which is
observed while measuring too high or too low resistances. We also studied about various
application of Wheatstone bridge method.
EXPERIMENT NO: 04
Name: Megha Mundra
DATE: 12-02-2021
Roll No. :19BEE056
AIM: - To measure low resistance using Kelvin's double bridge.
APPARATUS:1. Kelvin's double bridge
2. Unknown resistance
3. Galvanometer
4. Connecting wires
THEORY:Kelvins Double Bridge : This method is a modification of the Wheatstone bridge and provides greatly increased accuracy in
the measurement of low value of resistance. The Kelvin's double bridge incorporates the idea of
second set of ratio arms. Hence it is known as double bridge. It makes use of four terminal resistors
for the low resistance arms. The figure -1 shows the schematic diagram of the Kelvin’s bridge. The
first set of ratio arms is P and Q and the second set of ratio arms p and q is used to connect
the galvanometer to a point d at the appropriate potential between points m and n to eliminate
the effect of connecting lead of resistance r between the unknown resistance R and the standard
resistance S.
Replace delta connected resistances p,q,r into corresponding star connection.
𝑟1 = (
(𝑝∗𝑞)
p+q+r)
𝑟2 = (
𝑟3 = (
(𝑝∗𝑟)
p+q+r)
(𝑟∗𝑞)
p+q+r)
Now this connection replaces into simple Wheatstone bridge. Under balanced condition
𝑃
(R+r2)
𝑄
S+r3)
( )=(
𝑃∗(S+r3)
R + 𝑟2 =
R=
𝑄
𝑃∗(S+r3)
𝑄
− 𝑟2
qr
=
=
𝑃∗(S+p+q+r)
𝑄
𝑃∗𝑆
𝑄
+
−
pr
p+q+r
qr
𝑃
∗[
p+q+r
𝑄
𝑝
− ]-------------------- (1)
𝑞
If P /Q = p/q then,
R=
𝑃∗𝑆
𝑄
--------------------- (2)
Equation (2) is the usual working equation for the Kelvin Bridge. It indicates that the resistance of
connecting lead r has no effect on the measurement provided that the two sets of ratio arms have
equal ratios. Equation (1) is useful however as it shows the error that is introduced in case the ratio
is not exactly equal. It indicates that it is desirable to keep r as small as possible in order to
minimize the errors in case there is a difference between ratios P/Q and p/q. The effect of thermoelectric emfs can be eliminated by making other measurements with the battery connection reversed.
The true value of R is the mean of two readings.
PROCEDURE:1. Connect the circuit as shown in the figure.
2. Set suitable ratio of P/Q and p/q.
3. Obtain balance condition by the adjustment of standard resistance S.
4. Note down the value of the ratio of P/Q (or p/q) and S.
5. Reverse the polarity of the battery and repeat the above procedure.
6. Calculate the value of R for both conditions of the battery. True value of the resistance is the
mean value.
CIRCUIT DIAGRAM :
OBSERVATIONS:SR P/Q
STANDERD
UNKNOWN
THEORITICAL
NO RATIO RESISTANCE RESISTANCE UNKNOWN
SΩ
R = PS/Q Ω
RESISTANCE
Ω
ERROR
(%)
1
1/10
6.5
0.65
0.67
2.98
2
1/14
6.6
0.47
0.49
4.08
3
1/20
15.25
0.7625
0.8
4.6875
4
1/25
22
0.88
0.9
2.22
CONCLUSION: In this experiment we learned that a kelvin double bridge is a modified version of
the Wheatstone bridge, which can measure resistance values in the range between 1 to 0.00001
ohms with high accuracy.
LEARNING OUTCOME: We learned the principle behind the working of kelvin double bridge,
circuit diagram and its equivalent formula for finding the unknown resistance value, advantage and
application of this method. Lastly, we observed that while calculating the resistance through the
kelvin double bridge method we observed error in the range of 2-5 % because of certain
instrumental and human errors and also due to the error caused by thermoelectric emf which can be
eliminated by reversing the battery connection and taking the average of both the readings
QUIZ: 1. What are the advantages of Kelvin double bridge?
•
Errors owing to contact resistance, resistance of leads can be eliminated by using this
method.
•
•
•
It can measure the low resistance value.
Power consumption is less
Sensitivity is high.
2. What are the different difficulties encountered in measurement of low resistance?
Sol: The major problem in measurement of low resistance values is the contact resistance or
lead resistance of the measuring instruments, though being small in value is comparable to
the resistance being measured and hence causes serious error.
3. What are the limitations of VA method over Kelvin's double bridge?
Sol: The voltmeter ammeter method is a simple method but is not an accurate method of
measuring resistance. This method does not give high accuracy result because the error in the value
of resistance depends upon the accuracy of voltmeter and ammeter. An ammeter will have a small
internal shunt resistance which will sum with the wire resistance. If using a fixed voltage source
then the ammeter resistance will lower the circuit current. Whereas when we consider the kelvin
double bridge we get high accuracy result and also the errors owing to contact resistance, resistance
of leads can be eliminated by using this method.
EXPERIMEMNT NO : 05
NAME: MEGHA MUNDRA
DATE: 14-03-2021
ROLL NO.: 19BEE056
AIM: - To measure inductance by Anderson’s bridge and Hay's bridge
APPARATUS:1. Anderson’s bridge
2. Hay’s bridge
3. Galvanometer
4. Multimeter
THEORY: Anderson’s Bridge
The Anderson’s bridge is a modification of Maxwell’s bridge. Anderson’s bridge is used for
measurement of self inductance. This method is suitable for precise measurement of self inductance
over a very wide range of values.
Let
L1 = Unknown self inductance.
R1 = Series resistance of L1
C = Fixed standard capacitor.
At balance I1 = I3 and I2 = Ic + I4.
Now
I1 * R3 = Ic * ( 1/ jωC)
Hence Ic = I1 * jωCR3
Writing other balance equations
I1(R1+jωL1) = I2R2 + Icr
Ic(r + (1/jωc)) = (I2 – Ic) * R4
Equating real and imaginary parts
R1 = (R2 * R3/R4)
L1= C * (R3/R4) [r(R4+R2) + R2 * R4]
Advantage:1. Wide range of inductance can be measured
Disadvantage:1. The equations are complicated.
Hay's Bridge
The Hay’s bridge is a modification of Maxwell’s bridge. The connection diagram is shown in figure.
This bridge uses a resistance in series with the standard capacitor C unlike the Maxwell’s bridge
which uses a resistance in parallel with capacitor.
Let
Lx = Unknown inductance.
Rx = Series resistance of Lx
R1,R2,R4= Variable non-inductive resistor.
C1 = Variable standard capacitor
At balance (Rx + jωLx). (R1 – j / ωC1) ) = R2R4
Equating real and imaginary parts results
Rx = (ω2R2R1R4C12) / (1+ω2C12R12)
Lx = (R2R4C1) / (1+ω2C12R12)
The Q factor of the coil is given below.
Q = (ωLx)/Rx = 1 / (ωC1R1)
Therefore Lx = R2R4C1 / (1+ (1/Q2))
For Q > 10, the term (1/Q2) can be neglected.
Therefore Lx = R2R4C1.
Advantages:1. Gives very simple expression for unknown inductance for high Q coils and is suitable for
coil having Q>10.
2. The bridge also gives a simple experiment for Q factor.
3. This bridge requires a low value of resistor for R1 for high Q coil.
Disadvantage:1. This bridge is not suitable for measurement of coil having Q < 10.
PROCEDURE :Anderson’s Bridge
1.
Connect the unknown inductor in bridge as shown in fig. select any one standard capacitor
(known) from given capacitor bank Cx.
2.
Connect the detector.
3.
Switch on the source.
4.
Vary R1 to achieve null point on detector.
5.
Now put the mains switch off.
6.
Measure the value of r.
7.
Calculate the value of inductance Lx
Hay's Bridge
1.
Connect the unknown inductor in bridge as shown in fig.
2.
Select one of the standard capacitor using band switch.
3.
Switch on the sourc, adjust frequency to 1kHz.
4.
Set the value of Rx for minimum detector output.
5.
Vary R2 try to obtain null point on detector
6.
If null point is not achieved change value of Rx , C1 and frequency and
again repeat
step 4 nad 5.
7.
If null point is achieved switch off the mains.
8.
Measure the value of R2 .
9. Calculate the value of inductance L1 = R2R3C4
OBSERVATIONS:Anderson’s bridge
R3=R4=1 kΩ, R2=1.5 kΩ
Sr.
R5
Cx
Unknown Inductance
No
Ω
µF
L1= Cx * (R2/R3) [R5 * (R4+R3) +
R3*R4]
1.
147
0.02
38.82
2.
144
0.2
386.4
3.
18
2
3.108
Hay's Bridge
Sr. No
C1
R2
R4
Frequency
Standard value of
Unknown
nF
kΩ
kΩ
kHz
Inductance
Inductance
mH
L1 = R2*R4*C1
1.
150
0.39
0.173
10
11
10.12mH
2
100
0.39
1.219
10
50
47.54mH
3.
47
1.152
1.88
10
160
101.79mH
CONCLUSION: - In this experiment we studied the basic theory and equation of AC bridges then
we studied briefly about Maxwell’s bridge which is used for measurement of self-inductance. There
are two different bridges in Maxwell’s Bridge, one is Maxwell Inductance Bridge and second one is
Maxwell Inductance-Capacitance Bridge. We derived their equation and learnt to draw phasor
diagrams for each of the bridges and then briefly looked upon its advantages and disadvantages.
Since the maxwell bridge cannot be used for low Q factor we further studied about Hay’s bridge
which is a modification of Maxwell Inductance Capacitance Bridge then subsequently we derived
the expression and its phasor diagram and simulated the same using PSIM software .Finally we
studied Anderson’s bridge , its theory ,derivation of expressions and phasor diagram and verifying
our theoretical knowledge by simulating the AC bridge in PSIM software and learnt about the
advantages and disadvantages of Hay’s and Anderson’s bridge and their application.
CIRCUIT DIAGRAM :
QUIZ:1. Why the value of Q must be between 1 to 10 for Maxwell's bridge?
In the case of attempting to measure a high Q inductor one of the variable resistors must be a high
value resistance. It is expensive/impractical to get very accurate variable resistors of a high enough
value for use in a Maxwell bridge.
2. State the advantages and disadvantages of Maxwell’s bridge.
Advantages:
•
•
The frequency does not appear in the final expression of both equations,
hence it is independent of frequency.
Maxwell’s inductor capacitance bridge is very useful for the wide range of
measurement of inductor.
Disadvantages:
•
•
The variable standard capacitor is very expensive.
The bridge is limited to measurement of low-quality coils (1 < Q < 10) and it
is also unsuitable for low value of Q (i.e., Q < 1) from this we conclude that a
Maxwell bridge is used suitable only for medium Q coils.
3. In Maxwell’s bridge comparison___ can be made to read the value of unknown inductance
directly.
4. Give the application of Maxwell Inductance Capacitance Bridge.
In this method, the unknown inductance is measured by comparison with
standard known capacitance.
5. Hay’s bridge is modification of Maxwell Inductance Capacitance bridge.
6. State the advantages and disadvantages of Hay’s bridge.
Advantage of Hay’s Bridge:
•
It gives a simple expression for the unknown inductances and are suitable for coil
Having the quality factor greater than 10 times.
•
Simple expression of quality factor.
•
Uses small value resistance for determining the Q factor.
Disadvantages Of Hay’s Bridge:
•
Not suitable for the measurement of coil having quality factor less than 10 ohms.
EXPERIMENT NO: 06
NAME: MEGHA MUNDRA
DATE: 18-03-2021
ROLL NUMBER: 19BEE056
AIM: - To measure capacitance by De-sauty's bridge.
APPARATUS:1. De sauty’s bridge
2. Sine wave generator
3. Digital Multimeter
THEORY: De Sauty's Bridge:
De Sauty’s bridge is the simplest method of comparing two capacitance. The connection diagram is
shown in figure.
Let C1 = Unknown Capacitance
C2 = Standard Capacitor
R3, R4 = Variable non inductive resistor.
At balance (1/jωC1)R4 = (1/jωC2)R3
Equating real and imaginary parts results
C1 = C2 * (R4 / R3)
The balance condition can be obtained by varying R3 and R4.
The advantage of this bridge is its simplicity. But this advantage is nullified by the fact that it is
impossible to obtain balance if both capacitors are not free from dielectric loss. Thus with this method
only air capacitors can be compared.
PROCEDURE:De Sauty's Bridge
1. Select one of the standard capacitor C3 and connect it in the circuit as shown.
2. Select one of the unknown capacitor C2.
3. Connect the multimeter ass shown in the circuit.
4. Select appropriate value of R2 and adjust R1, such that null point is achieved.
5. Disconnect the circuit and note down th value of R1, R2 and C3.
6. Connect different value of unknown capacitor and repeat all the above steps.
OBSERVATIONS:De Sauty's Bridge:
Sr. No
R1
R2
C3
Unknown Capacitance
(Ω)
(Ω)
( uF )
C2= C3 * (R2/R1).
( uF )
1.
14
13
1
0.928
2.
20
10
2
1
3.
15
20
3
3.99
CONCLUSION:-In this experiment we studied briefly about the De sauty’s bridge which is most
suitable method for comparing the two values of capacitor .But this method gives accurate result
only when we neglect dielectric losses in the bridge circuit. So to overcome this limitation we
studied about the modification made by Grover and analysed the phasor diagram for both De
sauty’s bridge and modified De sauty’s bridge.
QUIZ:1. State the limitations of De-sauty’s bridge.
Its almost impossible to obtain balance if capacitors are not free from dielectric losses.
2. How you can modify De-sauty’s bridge to overcome these limitations?
To modify the De Sauty’s bridge so that it will gives accurate results for imperfect capacitors also
Grover has introduced a modified De Sauty bridge electrical resistances r1 and r2 above on arms 12 and 4-1 respectively are introduced in the circuit , in order to include the dielectric losses. Also,
he has connected resistances R1 and R2 respectively in the arms 1-2 and 4-1.Themodified circuit is
shown below .
3. What is meant by the ‘dissipation factor’ of a leakage capacitor?
Dissipation factor (tan δ)or DF is defined as the ratio of the ESR and capacitive reactance. It is the
Dissipation factor is also known as the tangent of the loss angle and is commonly expressed in
percent. For a good capacitor Dissipation factor is small. It will vary depending on the dielectric
material and the frequency of the electrical signals.
Formula of dissipation factor is DF=ESR/Xc or tan Φ.
CIRCUIT DIAGRAM:
EXPERIMENT NO :- 07
DATE:- 1-04-2021
Name :Megha Mundra
Roll No. : 19BEE056
AIM: - To test the performance of single phase induction type energy meter.
APPARATUS:1. Energy meter 0-1500 rev/KWH
1
0-300V
0-5-10Amps
2. Voltmeter(MI) 0-300 V
1
3. Ammeter (MI) 0-10 Amps
1
4. Lamp bank
1
5. Stop watch
1
THEORY: The single phase induction type energy meter consists of two coils namely current coil (CC)
and pressure coil (PC). The current coil is made of few turns of thick wire and is connected in
series with the load .The pressure coil is made of many turns of thin wire and is connected in
parallel with supply. The magnet M1 is excited by current coil and is known as series magnet.
Another magnet M2 is excited by the pressure coil and is known as shunt magnet. The
alternating flux Φ1 produced by the series coil is proportional to and in phase with the line
current. The alternating flux Φ2 produced by pressure coil is proportional to supply voltage (V)
and lags behind by 90o. The exact phase displacement of 90o is achieved by adjustment of
copper shading band C provided on the shunt magnet as shown in fig.1. Major portion of Φ2
crosses the narrow gap between the center and side limb of M2 and a small amount, which is
the useful flux, passes through the disc D. The two fluxes Φ1 and Φ2 induce emfs in the disc
which further produce the circulatory eddy currents. The reaction between these fluxes and
eddy currents produces the driving torque in the disk.
The pair of magnets, which are mounted diametrically opposite to the magnets M1 and M2,
provides the braking torque. When the peripheral portion of the rotating disk passes through
the air gap of braking magnet, the eddy currents are induced in it, which give rise to the
necessary torque. This torque is proportional to speed of the disc,
Hence Tb  N
-----------------------(1)
As the torque on the disc is proportional to both fluxes 1 and 2 and the angle between them,
So Td  1*2.sin 
But 1  Ip  Vs
and 2  IL
Hence Td  Vs. IL . sin 
 Vs.IL . cos ( 90 - )
 Vs.IL . cos 
i.e.
Td  power
( because  = 90 -  )
-----------------------(2)
The disc achieves a steady speed when,
Tb = Td
-----------------------(3)
From (1), (2) and (3)
N  power
Integrating both the sides with reference to time we get,
 N dt   power . dt
 No of revolution  Energy
Hence the total revolution made by the disc of energy meter in a given period of time is
proportional to the energy measured during that time period.
The register mechanism (generally of cyclometer type) is used to record the revolutions and it
is calibrated such that it would directly measure the energy (kWh)
•
A single phase energy meter may be tested by two methods
1. Substandard energy meter method and
2. Stop watch method
In the first method the measured value of the energy is compared with the true value of the
energy obtained from the substandard energy meter, for the same time period. In the second
method, the measured value of the energy is compared with the true value of the energy
calculated from the equation,
ET = V* I * t. kWh
1000 x 3600
Where t is the time for which energy is measured.
In both the cases error is calculated by,
Percentage error = [ ET - EM ] * 100
ET
PROCEDURE:1. Connect the circuit as shown in figure.
2. Apply the resistive load and record voltage, current and time taken for 15 revolutions.
3. Change the load and repeat the above procedure. Take five sets of reading and tabulate
them.
4. Plot the graph of load v/s percentage error.
CIRCUIT DIAGRAM:
OBSERVATION: SR.NO
SUPPLY
LOAD
TIME FOR 15
TRUE VALUE OF
MEASURED
VOLTAGE
CURRENT
REVOLUTIONS
ENERGY
ENERGY BY
Vs ( Volts)
IL (Amps)
t ( sec)
ET =
VS* IL * t .
1000 * 3600
ENERGY METER
EM = 15/N(kWh)
(kWh)
1
230
1.2
125
0.00958
0.01
2
230
2.7
63
0.01087
0.01
3
230
3.9
41
0.01021
0.01
4
230
4.8
35
0.01073
0.01
CALCULATION:Energy meter rating = __0-1500__ rev / kWh = N rev / kWh
EM = Energy measured for 15 revolutions = (15 / N) kWh
ET = True value of energy =
VS* IL * t . kWh
1000x3600
Percentage error = [ ET - EM ] * 100 = _8.0036_ %
ET
RESULT TABLE:SR.
LOAD
NO
CURRENT
% ERROR = [ET - EM ]
ET
IL (Amps)
1
1.2
-4.3841
2
2.7
8.0036
3
3.9
2.0568
4
4.8
6.8033
X 100
GRAPH:
IL v/s %Error
6
5
4
3
2
1
0
-4.3841
8.0036
2.0568
6.8033
CONCLUSION: - From this experiment we studied the performance of single-phase
induction type energy meter. It is used to calculate the energy consumed in kWh. It is an
integrating type of instrument because it doesn’t show the values instantly rather store and
display them afterwards. We analyzed the performance of the energy meter using three
concepts: whenever a moving magnet is placed above the non-magnetic plate, plate also
moves in the same direction as of the magnet, If a stationary magnet is placed above the
moving non magnetic plate, magnet stops the movement of non-magnetic plate through the
process of generating and motoring action using Fleming’s right-hand rule and Fleming’s
left-hand rule and a moving magnetic field is produced with the help of three electromagnets
out of which two are connected in series and the current in the third electromagnet lags by 90
degree with the current in the other two electrogmagnet connected in series. Then performing
the experiment practically, we calculated the percentage error in energy and plotted the graph
of load v/s %Error.
QUIZ: 1. The household energy meter is which type of instrument?
Ans: The housing energy meter is integrating type of instrument.
2. Why shading bands are used in energy meter?
Ans: The shading bands are used in energy meter due to the following reasons:
•
For phase error correction.
•
For elementary errors.
•
For flux compassion.
•
For adjusting lag angle (usually 90°, between magnetic field and supply voltage.)
3. What will be the speed of the disc for measured energy of 3.2 kW? The meter
constant is 1200 rev/ kWh.
Ans: Measured energy = 3.2 kWh
Meter constant = 1200 rev/ kWh
𝑟𝑒𝑣
Meter constant = 𝑘𝑊ℎ
𝑟𝑒𝑣
1200 = 3.2
𝑟𝑒𝑣 = 3840 𝑟𝑝𝑚