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Concordia University *
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ELEC 273
Subject
Electrical Engineering
Date
Jan 9, 2024
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LABORATORY REPORT
Basic Circuits & Systems Laboratory
This cover page must always be the top sheet
Course:
ELEC 273
275
Lab Section:
( Circle )
Experiment No.:
Date Performed:
20
–
–
YYYY
–
MM
–
DD
Experiment Title:
Name:
ID No.:
Lab Partner
Name:
Lab Partner
ID:
I certify that this submission is my original work and
meets the Faculty’s Expectations of Originality
Signature:
Date:
20
–
–
YYYY
–
MM
–
DD
FL-X
Prince Raphael Johnson
40153375
Faisal Quraishi
40161298
23
10
11
#4
Measurement of AC Power
23
11
20
Abstract
The objective of this experiment was for the students to familiarize themselves with the
measurement of AC power being delivered to a resistive load, an RC load, as well as an RL load.
This will be done using a Multi-Function Digital Meter, a Wattmeter as well as an AC Voltage
Source
Introduction
Just like previously, this experiment will revolve around AC measurements, in this case
AC power. The AC circuit will be the equivalent of a DC circuit, with the impedance being
equivalent to the resistance in this context (and measured in ohms) as well as the inductance and
the capacitance. The Digital meter used during the experiment will be reading the RMS voltage
across the various loads as well as the RMS current flowing through them.
Procedure (Methods)
For the various tasks, always 1
st
make sure that Variac knob is at its zero position and never leave
the R control knob at 0 to avoid making the load 0 ohms and possibly blowing the fuse in the load
unit when power is applied
Power Measurement
-
1
st
calibrate the rheostat in the RL load unit using the ohmmeter to get the resistance at
each dial
-
Proceed to connect the RL unit, have the R dial to 40 and slowly increase the Variac setting
until the meter reads about 1 A. From there, we get the value of the voltage, current, power
and power factor and increase R afterward to get the next values for the different Resistive
loads
-
Repeat the same process for RL and RC load but have the R dial to 10 1
st
. In the case of
the RC load, do not forget to disconnect the inductance of the RL load unit and set up the
RC load by connecting the resistor in the RL load unit in series with the C load unit.
Power Factor Compensation
-
Here, with the Variac again set to 0, we set up the circuit between the capacitor load unit
(with all the switches down at 1
st
) and the RC unit and have the R dial set to 30. Again,
slowly increase the Variac setting until the meter reads about 1 A
-
1
st
record the voltage, current and power with the capacitance set to
𝐶
= 0, then successively
switch in C values from 1 µF to 40 µF and proceed to record the voltage, current and power
values as
𝐶
is increased
Results and Discussion
Power Measurement
-
Does the measured value of the power factor agree with the theoretical value?
-
We have the theoretical
𝑝𝑓
= cos
𝜃
, with
𝜃
= tan-1 (
𝜔
L/R) for the RL load and
𝜃
= tan
-1
(-
1
ωC
𝑅
) for the RC load
o
At 60Hz, for the RL load,
𝜔
L= 2π60*0.204H
For R= 40,
𝜃
= 62.52°
/ For R=60,
𝜃
= 52.04°/ For R= 80,
𝜃
=43.87°
o
At 60Hz, for the RC load,
𝜔
C= 2π60*(40*10
-6
F)
For R= 10,
𝜃
= -81.425°
/ For R=30,
𝜃
= -65.659°
For R= 50,
𝜃
= -52.984°
-
The calculated pfs for the resistive load were not far from the measured power factor as
demonstrated by the % error, which was not the case for either the RL or RC load who’s
calculated pfs were very far from the measured ones. In that case, the calculated pfs for the
RL or RC load were closer to the measured pf of the resistive load actually.
R
Load
Measured
RMS
Voltage
𝑉
rms
Measured
RMS
Current
𝐼
rms
Measured
average
power
?
Measure
d
power
factor
𝑝𝑓
Calculated
𝑝𝑓
=
?
/
(𝑉
rms
*
𝐼
rms
)
Theoretical
Power factor
𝑝𝑓
= cos
𝜃
% error in
the
measured
power
factor
compared
to the
calculated
value
R=40
Ω
34
0.89
30.6 W
1.0
1.011
0
-1.1%
R=60
34.1
0.58
19.8
1.0
1.001
0
-0.1%
R=80
34.3
0.43
14.7
1.0
0.9966
0
0.34%
RL
load
204 mH
R=10
98.8
0.89 A
88.2
0.23
1.003
cos(62.52°)
=
0.461
-336%
R=30
98.7
0.83
81.9
0.39
0.99974
cos(52.04)
=
0.615
-179.3%
R=50
98.8
0.77
76
0.59
0.9990
cos(43.87)
=
0.721
-69.32%
RC
Load
40 µF
R=10
56.4
0.89
49.7
0.07
0.9901
cos(-81.425°)
=
0.149
-1314.4%
R=30
56.2
0.80
44.9
0.36
0.9987
cos(-65.659°)
=
0.412
-177.42%
R=50
56.3
0.69
38.9
0.55
1.001
cos(-52.984°)
=
0.602
-82%
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Load
Measured
Apparent
power
?
APP
=
𝑉
rms
𝐼
rms
(watts)
Measured
power
factor
𝑝𝑓
Measured
Angle
𝜃
=±cos
-1
(
𝑝𝑓
)
Theoretica
l
Angle
of
the
impedance
𝜃
Reactive
power
?
=
?
AP
sin
𝜃
(vars)
Complex
power
𝑆
=
?
+
𝑗?
%
error
in
angle
𝜃
RL
load
R=
40
P=98.8*0.89
=
87.932
0.23
+cos
-1
(0.23)
=
76.70°
= 62.52°
Q=
87.932sin
(76.70)
=
85.57
87.932+j85.57
18.49%
R=
60
98.7*0.83
=
81.921
0.39
=
67.05°
= 52.04°
=
75.44
81.921+j75.44
22.39%
R=
80
98.8*0.77
=
76.076
0.59
=
53.84°
= 43.87°
=
61.42
76.076+j61.42
18.52%
RC
load
R=
10
56.4*0.89
=
50.196
0.07
-cos
-1
(0.07)
=
85.99°
=
-81.425°
=
50.07
50.196
+
50.07
195%
R=
30
56.2*0.80
=
44.96
0.36
=
68.90°
=
-65.659°
=
41.95
44.96+j41.95
195%
R=
50
56.3*0.69
=
38.847
0.55
=
56.63°
=
-52.984°
=
32.44
38.847+j32.44
194%
-
As seen in the table, the measured angles for the impedances does not agree at all with the
theoretical values of
𝜃
for the RL load and
𝜃
for the RC load. For the RL Load, the
theoretical values were not completely far off, but that was not the case the RC load as the
theoretical values were all negative
Power Factor Correction
Measured
RMS
Voltage
𝑉
rms
Measured
RMS
Current
𝐼
rms
Measured
average
power
?
Measured
power
factor
𝑝𝑓
Calculated Power
Factor pf
% error
𝐶
= 0
𝜇𝐹
104.7 V
0.89 A
94.7 W
0.39
94.7/(104.7*0.89)
=
1.016
-160%
𝐶
= 5
𝜇𝐹
105.6
0.71
79.7
0.49
=
1.063
-117%
𝐶
= 10
𝜇𝐹
105.9
0.51
53.6
0.65
= 0.992
-53%
𝐶
= 15
𝜇𝐹
105.5
0.36
38.0
0.92
= 1.001
-9%
𝐶
= 20
𝜇𝐹
106.2
0.29
31.8
0.97
= 1.033
-6.5%
𝐶
= 25
𝜇𝐹
105.7
0.34
36.0
0.66
= 1.002
-52%
𝐶
= 30
𝜇𝐹
106.6
0.48
51.1
0.43
= 0.999
-132%
𝐶
= 35
𝜇𝐹
106.0
0.66
70.0
0.31
= 1.001
-223%
𝐶
= 40
𝜇𝐹
106.1
0.85
89.0
0.23
= 0.987
-329%
𝐶
for
minimum
current,
𝐶
=
20
𝜇𝐹
106.2
0.29
31.8
0.97
1.033
-6.5%
-
For each capacitance value, the power factor is calculated using
𝑝𝑓
=
?
/
(𝑉
rms
*
𝐼
rms
)
-
For
𝐶
= 0
𝜇𝐹
, the power factor is similar the measured one with the RL circuit for R=60
Ω
-
For minimum current, we used c=29.52 µF, R=33
Ω
, and in this case the current flowing
out of the generator using the LTSpice circuit is 0.584877A, which does not agree with the
measured current for similar value, but it is not far from it.
-
The power factor from the LTSpice simulation done in the prelab does not agree with the
one we measured during the lab itself. Does the value agree with your measurement?
Conclusion
Overall, the goal of this experiment was met, and it was useful for students to get familiar
with AC power measurement and calculations. Some of the results we got in the report varied
widely from the one measured during the lab, but it did show us that despite doing well during the
lab and every being correct, there will still be some differences when it comes to the report and the
actual theoretical calculations and value.
Related Questions
. Give the short answer offollowing questions each question carries equal marks
Don’t need tobriefly explain Its better to give point to point answer .
(a) How zenerdiode is a special diode
(b) Explain the difference between zener and avalanche effect
(c) Explain break down voltage
(d) Draw the symbol of PN junction diode, tunnel diode , Zener
Diode.
(e) How collector size is more than emitter
(f) How NPN transistor is more good as compare to pnp
(g) Why collector have more size than emitter
arrow_forward
Question 4
A zener regulation circuit is given in Figure 3.
R
100 0
VIN
VOUT
Figure 3
(a) Determine the minimum and maximum input voltages that can maintain the zener diode
IN4753A in the 'ON' state by using ideal zener diode.
1.
(b) Determine the minimum and maximum input voltages that can maintain the zener diode
IN4747A in the "ON" state by using practical zener diode.
arrow_forward
Please answer all subpart not only first three subpart I will like for all subpart either dislike is ready please..
arrow_forward
In a full-wave rectifier _______________.
a) the output is apure DC voltage.
b) only the negative half of the input cycle is used.
c) only the positive half of the input used.
d) the complee input cycle is used.
arrow_forward
What type of waveform is produced from a bridge rectifier circuit with an open diode?
a. Half-wave output equal to Vsec- 1.4
O b. Half-wave output equal to Vsec- 0.7
Oc. Full-wave output equal to Vsec- 1.4
Od. Full-wave output equal to Vsec- 0.7
QUESTION 14
To maintain a constant voltage supplied to the load, we can use
Oa. A high-resistance voltage divider
Ob. Either a Zener diode or an IC regulator
OC. A Zener diode
O d. An IC regulator
arrow_forward
The purpose of the current limiting resistor in a Zener diode circuit is to _________.
a. Maintain a constant current through the load resistor
b. Maintain a constant voltage across the Zener diode
c. Maintain a constant current through the Zener diode
d. Drop the additional voltage from the power supply in order to maintain a constant voltage across the Zener diode
Please answer it clearly.
arrow_forward
Discuss the different types of protections available for a Silicon controlled rectifier (SCR). In what kind of situations is it required to connect a number of SCRs in parallel and series respectively?
arrow_forward
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