Power Circuits and Transformers-Unit 2 Labvolt Student Manual

© Festo Didactic 30328-10 57
When you have completed this unit, you will be able to explain and demonstrate
the amplitude, frequency and phase of alternating voltage and current
waveforms. You will also demonstrate concepts related to instantaneous power.
Alternating current (ac) is universally used throughout the world for driving
motors and for powering electrical equipment. As its name suggests, an
alternating voltage is one which is continually reversing (alternating) its polarity.
When speaking of ac voltages, it is quite correct to consider them as being
dc voltages which are continually in the process of changing their value and
polarity. The number of times that the polarity passes from positive to negative
and then from negative to positive in one second is called the frequency. The
normal ac line frequency in North America is 60 Hz, while most countries in
Europe, and several others, have an ac line frequency of 50 Hz.
Besides reversing polarity periodically, ac voltages also change in value from
instant to instant, in a way that depends on the type of power supply. It is
possible to obtain a square wave, a triangular wave, or other types of waveforms
for the voltage. Theory and practical evidence has shown though, that the type of
waveform best suited for running electrical machinery is the sine wave. This
periodic waveform permits us to obtain the highest efficiency from transformers,
motors, and generators, and also results in the quietest operation. Although sine
waves seem more complicated than triangular or square waves, they make
calculations of voltages and currents in electrical circuitry simpler. The value of a
sine wave can be calculated for any instant of its cycle using the sine function,
and this value always repeats after one complete cycle.
Alternating Current
Unit 2
UNIT OBJECTIVE
DISCUSSION OF
FUNDAMENTALS

Unit 2 – Alternating Current  Discussion of Fundamentals
58 © Festo Didactic 30328-10
Figure 2-1 gives fractional values for a sine wave over a complete cycle. It can
be used to quickly calculate and sketch the waveform. Sine waves having
maximum values other than unity can be calculated using simple proportion.
Negative values indicate that the polarity of the voltage or current has reversed.
Figure 2-1. Sine wave values over a complete cycle.
Phase angle Relative amplitude Phase angle Relative amplitude
0° 0 180° 0
15° 0.26 195° -0.26
30° 0.50 210° -0.50
45° 0.71 225° -0.71
60° 0.87 240° -0.87
75° 0.97 250° -0.97
90° 1.00 270° -1.00
105° 0.97 285° -0.97
120° 0.87 300° -0.87
135° 0.71 315° -0.71
150° 0.50 330° -0.50
165° 0.26 345° -0.26
180° 0 360° 0

Unit 2 – Alternating Current  Discussion of Fundamentals
At any given instant in time, a sine wave will be at a given position, measured in
degrees from a reference point. Consider two identical generators adjusted to
exactly the same frequency. Suppose now that the second generator is turned on
a short instant after the first. When both waveforms are observed together on an
oscilloscope, the display will be similar to Figure 2-2. Using the sine wave from
the first generator as a reference, we can say that the second waveform is
lagging the reference by several degrees. The separation in time between the
two ac waveforms is the phase shift. Phase shifts are often measured using
phase angles. The term lagging or leading phase shift is used to indicate
whether the waveform reaches maximum after or before the reference.
Figure 2-2. Shift angle between two sine waves of identical frequency.

When you have completed this exercise, you will be able to measure the
amplitude and frequency of ac waveforms, and demonstrate concepts associated
with these parameters.
The sine wave has a direct relationship to circular rotation, as Figure 2-3 shows.
Each cycle of a sine wave is equivalent to one complete revolution, which
equals 360°. In fact, standard alternating voltage produced at the local power
plant is a sine wave. The voltage starts at zero, and then increases to a
maximum value. It then decreases until it reaches zero again, at which point the
voltage changes polarity. It then increases until a maximum at the opposite
polarity is reached, and once again it decreases towards zero. At the point where
the voltage reaches zero a second time, a full revolution of 360 angular degrees
has been completed. For a 60-Hz system, this means that in one second,
60 complete cycles of the sine wave take place. Therefore, the period of a
60-Hz sine wave is 1 60⁄ seconds.
The Sine Wave
Exercise 2-1
EXERCISE OBJECTIVE
DISCUSSION

Ex. 2-1 – The Sine Wave  Discussion
Figure 2-3. One full cycle of a sine wave equals ° of rotation.

Ex. 2-1 – The Sine Wave  Discussion
Amplitude and frequency are important parameters of a sine wave. The
amplitude can be indicated as a peak-to-peak value, peak value or rms value.
The maximum value reached by the sine wave during its cycle is the peak value,
while the peak-to-peak amplitude is simply twice the peak value. The rms
(root-mean-square) value or effective value is defined by the heating effect of
the sine wave. For example, suppose that a sine-wave voltage (ac voltage) with
a peak value of 100 V is connected to a load resistor and the resistor
temperature is measured once it has stabilized. The effective value of the
ac voltage can be found by using a variable dc supply, and adjusting the
dc voltage until the temperature of the resistor stabilizes at the same point as
before. The resulting dc voltage is 71 V, meaning that the rms value of the
ac voltage is 71 V. Another way to obtain a measure of the rms value consists in
applying a dc voltage to one lamp and an ac voltage to another. The lamp
brightness is a fairly accurate indicator of the power being dissipated, and the
dc voltage can be adjusted to obtain the same brightness as the ac voltage.
Naturally, these methods are time consuming and not very efficient for
determining the rms value of an alternating voltage or current.
Measuring instruments for standard (sine wave) alternating current are calibrated
to indicate the rms value directly. The rms value is related to the peak value by a
simple relationship, 0.707 (the peak value multiplied by 1 √2⁄ ).
Note that this relationship is only valid for sine waves. Also, the rms subscript is
usually indicated only if necessary.
Finally, the other important parameter of the sine wave, its frequency, is just the
reciprocal of the waveform's period, i.e., 1⁄ . For 60-Hz ac power systems,
the period is 1 60⁄ 0.0167 s and the reciprocal is 60 Hz. For 50-Hz ac power
systems, the period is 1 50⁄ 0.02 s and the reciprocal is 50 Hz. Conversely, the
period is the reciprocal of the frequency, 1⁄ . Figure 2-4 shows the
parameters of a sine wave.
Figure 2-4. Amplitude and period of a sine wave.

Ex. 2-1 – The Sine Wave  Procedure
EQUIPMENT REQUIRED
Refer to the Equipment Utilization Chart in Appendix C to obtain the list of
equipment required for this exercise.
High voltages are present in this laboratory exercise. Do not make or modify any
banana jack connections with the power on unless otherwise specified.
1. Install the Power Supply, data acquisition module, and Resistive Load
module in the EMS Workstation.
2. Make sure that the main power switch of the Power Supply is set to the
O (OFF) position, and the voltage control knob is turned fully
counterclockwise. Make sure that the Power Supply is connected to a three-
phase wall receptacle.
3. Set up the circuit shown in Figure 2-5. Note the symbol used to indicate a
variable ac source in this circuit. Set the Resistive Load module for the given
resistance value, and connect inputs E1 and I1 to measure the circuit voltage
and current.
Figure 2-5. AC sine wave circuit.
PROCEDURE
Local ac power network
(Ω)
Voltage
(V)
Frequency
(Hz)
120 60 300

4. Make sure that the data acquisition module is connected to a USB port of the
computer.
Connect the POWER INPUT of the data acquisition module to the
24 V - AC output of the Power Supply. Set the 24 V - AC power switch to the
I (ON) position.
5. Start the Data Acquisition software (LVDAC or LVDAM). Open setup
configuration file ES12-1.dai.
a The metering setup configuration can be changed during the exercise if
desired. This exercise was written using those given.
Make sure that the continuous refresh mode is selected.
6. Turn on the main Power Supply and adjust the voltage control knob to 100%.
7. Use the Data Table to store the values of the measured circuit voltage and
current.
8. Click on the Oscilloscope button and display E1 and I1 on CH1 and CH2. If
necessary, readjust the time base to show at least two cycles of the sine
waves.
9. Set convenient vertical scales for the display and note the peak amplitudes of
the voltage and current.
V
A

10. Multiply the peak values by 0.707 and compare with the results stored in the
Data Table.
0.707 V
0.707 A
11. Is there a difference between your calculations and the stored values?
a The results in the Data Table are rms values. Also, the waveform data box of
the Oscilloscope screen gives the rms value of the signals applied to the input
channels, their average value, and the frequency.
12. Compare the current waveform with the voltage waveform. Are they both
sine waves?
 Yes  No
13. What is the approximate amplitude of the voltage sine wave at 225°?
Amplitude V
14. What is the time period for one complete cycle of the ac voltage waveform?
ms

15. Calculate the frequency.
1
Hz
16. Compare the frequency of the current waveform to that of the voltage. Are
they the same?
 Yes  No
17. Do the current and voltage waveforms reach their maximum and minimum
values at the same time, as well as passing through zero amplitude?
a When the conditions of going through zero and reaching maximum at the
same time occur, the waveforms are said to be in phase, meaning that there is
no phase angle separation between them.
 Yes  No

18. Turn off the Power Supply and set up the series circuit in Figure 2-6. Set the
Resistive Load module for the given resistor values, and connect
inputs I1, E1, and E2 as shown in the figure.
Figure 2-6. AC series circuit.
19. Set the voltmeter select switch to the 4-N position. Turn on the Power Supply
and adjust the voltage control knob to obtain the value of voltage given in
Figure 2-6. Use the ES12-2.dai configuration file for the circuit
measurements.
20. Click on the Oscilloscope button and display E1, E2, I1 on CH1, CH2,
and CH3. Make sure that the time base control is set to show at least two
cycles of the sine waves.
21. Select convenient vertical scales for the display and note the rms values of
the voltages and the current.
E1 V
E2 V
I1 A
(V) (Ω) (Ω)Voltage
(V)
Frequency
(Hz)
120 60 120 171 300

22. Compare the ratio of the voltages to the ratio of the resistors.

23. Calculate the voltage drops for each resistor using the values given in
Figure 2-6 and compare with the measured rms values of step 21.
V
V
24. Calculate the value of source current that should flow in this circuit and
compare with the measured rms value of step 21.

A

25. Do the results of steps 21, 22, 23, and 24 demonstrate that Ohm's law,
Kirchhoff's laws, and the other circuit theory of Unit 1 is valid for ac circuits?
 Yes  No
26. Observe the current and voltage waveforms and notice that they have the
same type of relationship as the waveforms in step 17. Does this mean that
they are in phase?
 Yes  No
27. Ensure that the Power Supply is turned off, and that the voltage control knob
is turned fully counterclockwise. Remove all leads and cables.

Ex. 2-1 – The Sine Wave  Conclusion
In this exercise, you demonstrated that a sine wave of voltage produces a sine
wave of current, and that the rms value of a sine wave equals 0.707 times the
peak amplitude. You also confirmed that the frequency is the reciprocal of the
waveform's period, and you saw that the theory presented in Unit 1 is valid for
ac circuits.
1. The peak-to-peak amplitude of a sine wave is 200 V. What is the rms value?
a. 282 V
b. 70.7 V
c. 141 V
d. 14.1 V
2. The period of a sine wave is 0.02 seconds. What is its frequency?
a. 5 Hz
b. 50 Hz
c. 50 s
d. 0.02 Hz
3. An ac voltage can be considered as a dc voltage that is continually changing
its amplitude and polarity.
a. False.
b. True in cases when the current is zero.
c. True.
d. None of the above.
4. One complete cycle of a sine wave is the same as one circular rotation of
360°.
a. True in cases where the frequency is less than 100 Hz.
b. True.
c. False because a sine wave is not a circle.
d. False.
CONCLUSION
REVIEW QUESTIONS

Ex. 2-1 – The Sine Wave  Review Questions
5. When are two sine waves said to be in phase?
a. When the current leads the voltage.
b. When they both attain their maximum values at the same
time.
c. When they both go through zero at the same time.
d. Both b and c together.

When you have completed this exercise, you will be able to determine the phase
angle between sine waves, and distinguish between leading and lagging phase
shift.
Phase angle is used to measure the amount of separation (phase shift) between
two sine waves of the same frequency. The sine waves being compared must
have the same frequency, but they do not need to have the same amplitude. In
later units, you will see that components like capacitors and inductors cause a
phase shift between the voltage and current. The amount of phase shift between
two sine waves is often expressed as a phase angle. One of the two sine waves
is used as the reference for phase shift measurements.
To determine the phase angle using an oscilloscope, the reference waveform is
applied to one channel input, and the other channel receives the waveform for
which we want to measure the relative phase difference. Figure 2-7 gives an
example of how this can be done. The oscilloscope is adjusted so that one
complete cycle of the reference waveform (360°) is displayed over an exact
number of divisions, i.e., 8 divisions in this example. Each division is therefore
equal to 45°, and minor divisions correspond to 9°. Finally, the horizontal
separation (d) between the waveforms is measured, which in this example gives
0.8 division. The phase angle is therefore 0.8 45° 36°.
Figure 2-7. Phase angle between two waveforms.
Phase Angle
Exercise 2-2
EXERCISE OBJECTIVE
DISCUSSION

Ex. 2-2 – Phase Angle  Procedure
The second waveform lags the reference waveform since it reaches maximum
amplitude after the reference waveform. As the figure shows, a lagging waveform
is shifted to the right of the reference on the oscilloscope display. Since the
phase is lagging, it is common to see a minus sign or the word lagging included
with the number, i.e. 36°, or 36° lagging. This is a standard shorthand way of
indicating whether the phase is leading or lagging. If the second waveform were
shifted to the left on the oscilloscope, the phase shift would be leading, since the
second waveform reaches maximum before the reference waveform. Phase shift
could also be indicated by using positive numbers for phase differences greater
than 180°. If you examine closely the waveforms in Figure 2-3, it becomes clear
that a phase shift of 270° leading is the same as 90° lagging. Further along in the
study program, you will set up circuits with capacitors and inductors that cause
large phase shifts between voltages and currents.
EQUIPMENT REQUIRED
1. Install the Power Supply and data acquisition module in the
EMS Workstation.
3. Set up the circuit shown in Figure 2-8. Connect inputs E1, E2, and E3 to
measure the phase voltages of the Power Supply at 1-N, 2-N, and 3-N.
Make sure that that the data acquisition module is connected to a USB port
of the computer. Connect the POWER INPUT of the data acquisition module
to the 24 V - AC output of the Power Supply. Set the 24 V - AC power switch
to the I (ON) position.
PROCEDURE

Figure 2-8. Phase angle measurement.
 Yes
5. Turn on the main Power Supply and verify that the phase voltages are
displayed on the Metering application.
6. Click on the Oscilloscope button and display E1, E2, and E3 (that is,
voltages 1-N, 2-N and 3-N) on CH1, CH2, and CH3. Make sure that the time
base control is adjusted to show at least two complete cycles of the sine
waves.
7. If necessary, select convenient vertical scales for the amplitudes and use E1
as the reference waveform for phase shift measurement.
8. Looking at the three waveforms, is there a phase shift between them?
 No

9. How many degrees separate the voltage on
E1 from that on E2? °
10. Is the voltage on E1 leading or lagging the voltage on E2 by 120°?
11. Is the voltage on E3 leading or lagging the voltage on E1 by 240°?
12. You will have noticed that the voltages on E1, E2, and E3 are separated
by 120°, which is the case for standard three-phase alternating current. If the
voltage on E2 is now used as the reference waveform, is it leading or lagging
the voltage on E1 by 120°?
13. Change the time base setting on the oscilloscope to increase the number of
cycles displayed. Does the amount of phase shift between the waveforms
change?
 Yes  No

14. Change the vertical scale settings on the oscilloscope. Does the amount of
phase shift between the waveforms change?
 Yes  No
15. Turn on the Cursors, and use the left and right cursors to measure the time
difference between the waveforms displayed on channels E1, E2, and E3
at the point where they pass through zero amplitude.
(E1 - E2)
(E3 - E1)
(E2 - E3)
16. Determine the phase angle between the waveforms. Note that is the period
of the reference waveform.
Phase angle 360° °
17. Are the measured values similar to the results of step 9?
 Yes  No

Ex. 2-2 – Phase Angle  Conclusion
In this exercise, you observed three sine waves that were separated in phase by
120°. You also saw that to determine whether one waveform leads or lags
another, it is necessary to decide which waveform will be used as the reference.
1. Phase angle can be used as
a. a measurement of the period of a periodic waveform.
b. an indication of a signal's frequency.
c. a measurement of the separation between two waveforms.
d. only valid when three-phase signals are considered.
2. A sine wave has a leading phase angle of 72°. Will it reach maximum before
or after the reference waveform?
a. After.
b. Before.
c. It depends on the frequency.
3. Three-phase ac power consists of three sine waves separated by 120°?
a. True in North America only.
b. False.
c. True.
d. False since square waves are sometimes used.
4. A sine wave has a phase angle of 45°. Is the reference waveform leading
or lagging this sine wave?
a. Leading.
b. Lagging.
c. Neither, it is in phase.
d. The reference cannot lead or lag another waveform.
CONCLUSION
REVIEW QUESTIONS

Ex. 2-2 – Phase Angle  Review Questions
5. The oscilloscope waveforms of the current and voltage in a circuit show that
a large phase difference exists between the two. What does this indicate
about the type of circuit components?
a. Nothing
b. They must be defective.
c. They are all resistors.
d. There must be capacitors and/or inductors in the circuit.

When you have completed this exercise, you will be able to explain and
demonstrate the concept of instantaneous power. You will also be able to
determine the average power dissipated in a resistive load when it is connected
to a source of alternating current.
When electrical power is supplied to a dc motor, part of the power is converted
into mechanical energy and the remainder is converted into heat. When power is
supplied to a storage battery during charging, some of the power is converted to
chemical energy, while the rest is converted into heat. However, when power is
supplied to a resistor, all of it is converted into heat. This conversion of electrical
power into heat is a very efficient process, and we make use of it every day in
electric toasters, stoves, and electrically-heated homes.
As you have seen in Unit 1, power is the product of the voltage and the current in
dc circuits. The same is true for ac circuits, except that it is important to know
whether one is talking about instantaneous power or average power. From what
has been seen so far, it should be clear that the power dissipated by a resistor
connected to an ac source varies sinusoidally with time, since the circuit voltage
and current are sine waves. Instantaneous power is simply the product of
calculated at each instant in the sine wave cycle. If a wattmeter were connected
to measure power in the circuit of Figure 2-9, it would indicate a value other than
zero, even though the average value of the voltage and current waveforms is
zero over a period.
Instantaneous Power
Exercise 2-3
EXERCISE OBJECTIVE
DISCUSSION

Ex. 2-3 – Instantaneous Power  Procedure
This is what is shown by the instantaneous power waveform in Figure 2-9. Close
examination of this figure shows that when the voltage is negative, the current is
negative, so that the product is always positive. As you will learn in this
exercise, the average power dissipated by a resistive load is simply the product
of the rms voltage and current in the circuit. Another important point to note is
that the frequency of the instantaneous power waveform is twice that of the
source. This is because the product gives a sine-squared function which is
at twice the frequency of the original sine waves.
Figure 2-9. Instantaneous power waveform for a resistive load.
EQUIPMENT REQUIRED
1. Install the Power Supply, data acquisition module, and Resistive Load
module in the EMS Workstation.
PROCEDURE

3. Set up the circuit shown in Figure 2-10. Set the Resistive Load module for
the given resistance value, and connect inputs E1 and I1 to measure the
circuit current and voltage.
Make sure that the data acquisition module is connected to a USB port of the
computer. Connect the POWER INPUT of the data acquisition module to the
24 V - AC output of the Power Supply. Set the 24 V - AC power switch to the
I (ON) position.
Figure 2-10. Instantaneous power in a resistive load.
5. Turn on the main Power Supply and adjust the voltage control knob to 100%.
Verify that the circuit parameters are displayed on the Metering application.
6. Click on the Oscilloscope button and display E1, I1, and P1 on CH1, CH2,
and CH3. Make sure that the time base control is adjusted to show at least
two complete cycles of the sine waves.
(Ω)
Voltage
(V)
Frequency
(Hz)
120 60 171

7. Select convenient vertical scales for the display and note the rms values of
the voltage and current, and the average power (AVG) for P1.
E1 V
I1 A
P1 W
8. Compare the current waveform with the voltage waveform. Do they have the
same frequency?
 Yes  No
9. What are the time period and frequency of the instantaneous power
waveform?
ms
1
Hz
10. How does the frequency of the instantaneous power waveform compare with
those of the current and voltage waveforms?

11. Are the current, voltage, and power waveforms in phase?
12. Calculate the product of the rms values of the current and voltage and
compare it with the AVG value for P1 given in the waveform data box of the
Oscilloscope screen.
W
P1 W
13. Are the results approximately the same?
 Yes  No

14. Turn off the Power Supply and set up the circuit shown in Figure 2-11. Set
the Resistive Load module for the given resistance values, and connect
inputs I1, I2, I3, and E1, E2, E3 as shown in the figure.
Figure 2-11. Instantaneous power in a parallel circuit.
15. Open configuration file ES12-5.dai.
16. Turn on the Power Supply, adjust the voltage control knob to 100%, and
verify that the circuit parameters are displayed on the Metering application.
17. Click on the Oscilloscope button and display I1, I2, I3, and E1 on CH1, CH2,
CH3, and CH4. Make sure that the time base control is adjusted to show at
least two complete cycles of the sine waves.
(Ω) (Ω) (Ω)
Voltage
(V)
Frequency
(Hz)
120 60 171 171 171

18. Select convenient vertical scales for the display and note the rms value of the
voltage and currents below.
E1 V
I1 A
I2 A
I3 A
19. Calculate the product of the voltage and the currents to obtain the power
dissipated in each of the three resistors R1, R2, and R3.
W
W
W
20. What is the total power dissipated by the circuit?
W

21. On the Oscilloscope, display P1, P2, and P3 on CH5, CH6, and CH7. Note
the values given in the waveform data box.
P1 W
P2 W
P3 W
22. What is the total measured power?
W
23. Compare the results of steps 20 and 22. Is the total power approximately the
same in both cases?
 Yes  No
24. Compare the phase angles between the different waveforms. Is there any
appreciable phase shift?
 Yes  No
25. Do the instantaneous power waveforms of , , and confirm that the
power dissipated in a resistive circuit is always positive?
 Yes  No

Ex. 2-3 – Instantaneous Power  Conclusion
In this exercise, you demonstrated that the instantaneous power waveform
produced when alternating current is connected to a resistive load is always
positive and has a frequency twice that of the ac source. You also saw that the
average power dissipated by the load is the product of the rms values of the
circuit currents and voltages.
1. The average power dissipated in an ac circuit is equal to
a. zero over a period.
b. the square of the voltage divided by the current.
c. one-half the peak value.
d. equal to the product of the rms values of and .
2. The power waveform in an ac circuit is a sine wave at double the source
frequency.
a. True.
b. False, because the average power is greater than zero.
c. True, but only with a resistive load.
d. False, the power waveform is a square wave.
3. The instantaneous power waveform shows that the power is always positive
even though the voltage and current waveforms in an ac circuit alternate
between positive and negative values.
a. False.
b. True when the load is resistive.
c. True all the time.
d. It depends on the waveform frequency.
CONCLUSION
REVIEW QUESTIONS

Ex. 2-3 – Instantaneous Power  Review Questions
4. What is the average power dissipated by a 100-Ω resistor when it is
connected to an ac source with a peak value of 141 V
a. 1410 W
b. 14.1 W
c. 141 W
d. 100 W
5. Is it possible to have "negative power" in an ac circuit?
a. Yes, when capacitors or inductors are connected.
b. No.
c. Only if the source frequency is very low.
d. Negative power is possible only in extreme cases.

Unit 2 – Alternating Current  Unit Test
Unit Test
1. The rms value of a sine wave voltage equals 1 √2⁄ , which is the
dc equivalent value that will produce the same heating effect.
a. False.
b. True.
c. It depends on the load.
d. It depends on the frequency.
2. What is the period of a sine wave whose frequency is 120 Hz?
a. 120 ms
b. 8.3 ms
c. 33.4 ms
d. 16.7 ms
3. Alternating current is one which periodically changes direction and alternates
between maximum positive and negative values.
a. True.
b. False.
c. Only if the rms value is greater than zero.
4. What is the normal phase shift between the voltage and current in a circuit
with only a resistive load?
a. 180°
b. 90°
c. There is no phase shift.
d. It depends on the power dissipated in the load.

5. What is the period of a sine wave whose frequency is 50 Hz?
a. 35 ms
b. 25 ms
c. 20 ms
d. 16.7 ms
6. Knowing the phase angle between the voltage and current allows us to
a. know the frequency.
b. determine if the power is instantaneous.
c. determine the rms amplitude of the waveforms.
d. know if only resistors are present in the circuit.
7. Alternating current will produce more power in a given circuit than a
dc voltage with the same value as the peak ac voltage.
a. True.
b. False, because the average ac value is zero over a complete period.
c. False, because the dc voltage is greater than the rms ac voltage.
d. True, because power equals the product of the rms values of and .
8. The instantaneous power waveform in a dc circuit is the same as in an
ac circuit.
a. True.
b. False, it is a straight line.
c. True, but the frequency is very low.
d. False, instantaneous power does not exist in dc circuits.
9. It is impossible to solve ac circuits since Ohm's law and Kirchhoff's laws used
in solving dc circuits do not apply.
a. True.
b. False except when dealing with inductive loads.
c. True if the circuit only has resistors.
d. False, all the laws used so far apply to ac circuits.

10. What is the peak ac voltage required to produce the same average power as
a dc voltage of 50 V?
a. 141 V
b. 70.7 V
c. 50 V
d. 50 √2⁄ V

The following table gives impedance values which can be obtained using either
the Resistive Load, Model 8311, the Inductive Load, Model 8321, or the
Capacitive Load, Model 8331. Figure B-1 shows the load elements and
connections. Other parallel combinations can be used to obtain the same
impedance values listed.
Table B-1. Impedance table for the load modules.
Impedance (Ω) Position of the switches
120 V
60 Hz
220 V
50 Hz/60 Hz
240 V
50 Hz
1 2 3 4 5 6 7 8 9
1200 4400 4800 I
600 2200 2400 I
300 1100 1200 I
400 1467 1600 I I
240 880 960 I I
200 733 800 I I
171 629 686 I I I
150 550 600 I I I I
133 489 533 I I I I
120 440 480 I I I
109 400 436 I I I I
100 367 400 I I I I I
92 338 369 I I I I I
86 314 343 I I I I I I
80 293 320 I I I I I I I
75 275 300 I I I I I I I
71 259 282 I I I I I I
67 244 267 I I I I I I I
63 232 253 I I I I I I I I
60 220 240 I I I I I I I I
57 210 229 I I I I I I I I I
Impedance Table for the Load Modules
Appendix B

Appendix B Impedance Table for the Load Modules
Figure B-1. Location of the load elements on the Resistive Load, Inductive Load, and Capacitive
Load, Models 8311, 8321, and 8331, respectively.

Appendix B Impedance Table for the Load Modules
The following table gives inductance values which can be obtained using the
Inductive Load module, Model 8321. Figure B-1 shows the load elements and
connections. Other parallel combinations can be used to obtain the same
inductance values listed.
Table B-2. Inductance table for the Inductive Load module.
Inductance (H) Position of the switches
120 V
60 Hz
220 V
50 Hz
220 V
60 Hz
240 V
50 Hz
1 2 3 4 5 6 7 8 9
3.20 14.00 11.70 15.30 I
1.60 7.00 5.80 7.60 I
1.07 4.67 3.88 5.08 I I
0.80 3.50 2.90 3.80 I
0.64 2.80 2.32 3.04 I I
0.53 2.33 1.93 2.53 I I
0.46 2.00 1.66 2.17 I I I
0.40 1.75 1.45 1.90 I I I I
0.36 1.56 1.29 1.69 I I I I
0.32 1.40 1.16 1.52 I I I
0.29 1.27 1.06 1.38 I I I I
0.27 1.17 0.97 1.27 I I I I I
0.25 1.08 0.89 1.17 I I I I I
0.23 1.00 0.83 1.09 I I I I I I
0.21 0.93 0.77 1.01 I I I I I I I
0.20 0.88 0.73 0.95 I I I I I I I
0.19 0.82 0.68 0.89 I I I I I I
0.18 0.78 0.65 0.85 I I I I I I I
0.17 0.74 0.61 0.80 I I I I I I I I
0.16 0.70 0.58 0.76 I I I I I I I I
0.15 0.67 0.55 0.72 I I I I I I I I I

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Power Circuits and Transformers-Unit 2 Labvolt Student Manual