7labo ganago student lab8
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Lab 8: AC Power
LAB EXPERIMENTS USINGNI ELVIS II
AND NI MULTISIM
Alexander Ganago
Jason Lee Sleight
University of Michigan
Ann Arbor
Lab 8AC Power
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Lab 8: AC Power
Goals for Lab 8
Learn about:
The effects of phase shift between the input (source) voltage and the currentthrough the circuit on the transfer of power from the source to the load Calculations of the average power absorbed by the load, and the importance of the
Power Factor
Compensation of the unwanted phase shift between the source voltage and thecurrent in the circuit by way of adding a compensating impedance such as a
compensating capacitor
The frequency dependence of the compensation of the phase shift and of theoptimization of the power transfer to the load.
Learn how to:
Simulate various circuits in order to study the effectiveness of power transferfrom the source to the load
Relate your own sensory experience (hearing the sounds of varied loudness atvarious frequencies) to the measurements of voltages and calculations of the
power transferred to the load
Apply the concepts of Thevenin equivalent circuit to the power transfer from anAC source to the load
Measure the delays between the source voltage and the current through the circuitin several circuits, at various frequencies, in order to assess and optimize the
power transfer
Explain your lab data using several circuit models (for example, taking intoaccount the DC resistance of the inductor)
Compare the effectiveness and reliability of several circuit models in explainingyour lab data
Optional goals:
Explore the effects of varied compensating capacitance on the optimization of thepower transfer to the load at various frequencies
Learn about the importance of using a transformer; explore what exactly changesin your circuit if the transformer is removed.
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Lab 8: AC Power
Introduction
You have already learned that, in a linear circuit with DC sources and resistors, which is
shown as its Thevenin equivalent in Figure 8-1 (to the left of the terminals a and b), the
maximal transfer of power from the source to the load (to the right of the terminalsa
andbin Figure 8-1) is achieved when the load resistance equals (matches) the equivalent
resistance of the circuit:
Load TR R=
and the maximal power transferred to the load equals:
( )2
T
MAX
T
V1P
4 R= .
Figure 8-1. The Thevenin equivalent of a linear circuit with DC sources and resistors (tothe left of the terminals a and b) and the load resistor (to the right of the terminals a and
b).
When the DC (time-independent) sources are replaced with AC (time-dependent)
sources, we have to consider more aspects of the power transfer problem.
In particular, when all sources in the circuit are sinusoidal and at the same frequency (see the Thevenin equivalent circuit in Figure 8-2), we can calculate voltages, currents,
and power using either trigonometric equations with time-dependent voltages and
currents, or complex algebra with phasors.
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Lab 8: AC Power
Figure 8-2. The Thevenin equivalent of a linear circuit with AC sources and impedances
(to the left of the terminals a and b) and the load impedance (to the right of the terminalsa and b).
If we assume that the Thevenin equivalent voltage is the input signal, which serves as the
reference with zero phase angle, we can write:
T T, pk V (t) V cos( t)=
or, in the phasor form:
T T, maxV 0= V
The voltage across the load is also sinusoidal at the same frequency but, in general, it has
a different phase angle:
Load Load, pk V, LoadV (t) V cos( t ) = +
or, in the phasor form:
Load Load, max V, LoadV = V
the current I in the circuit is also sinusoidal at the same frequency but, in general, it has adifferent phase angle:
k II(t) I cos( t ) = +
or, in the phasor form:
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Lab 8: AC Power
k II = I
The average power absorbed by the load depends not only on the peak values of thevoltage and current but also on the phase shift:
av Load, pk pk V, Load I
1P V I cos(
2) =
or, in the phasor form:
( )av Load, pk pk V, Load I1
P V I2
=
If the phase angle shift ( )V I between the voltage and current equals zero, the power
reaches its maximum (for the given peak values of voltage and current).
On the contrary, if the phase shift ( )V I 902
= = , the average power equals
zero, regardless of how large the amplitudes of voltage kV and current kI could be.
This is the case of an individual inductor or capacitor.
The cosine of ( )V I has a special name: the Power Factor:
( )V Icos Power Factor =
By definition, the Power Factor ranges from 0 to 1, reaching exactly zero for an
individual inductor or capacitor, and reaching exactly 1 when the circuit or load is purelyresistive.
In power transfer, the goal is to maximize the Power Factor by minimizing ( )V I .Since many loads (such as motors) are inductive, and transmission lines also incur
inductance, a compensating capacitor can be added to the circuit in order to reduce
( )V I .
Note an important distinction between DC circuits and AC circuits with sinusoidalsources: A transmission line in a DC circuit adds extra resistance (see Figure 8-3), and
adding another resistor does not improve the power transfer to the load.
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Lab 8: AC Power
Figure 8-3. DC circuit with a transmission line that has its resistance RBLineB added to the
circuit.
A transmission line added to an AC circuit also contributes to the circuit resistance;
moreover, it adds some inductance to the circuit (Figure 8-4).
Figure 8-4. AC circuit with a transmission line has its impedance added to the
circuit.
LineZ
The line impedance is often inductive:LineZ
Line Line LineZ R j L= +
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Lab 8: AC Power
It may seem counterintuitive, but you will see it in your own experiments (not only in
calculations and simulations) that, in order to maximize the power transfer to the load, itis worthwhile to add an extra impedance to the circuit: a properly chosen compensating
capacitor reduces the overall phase shift ( )V I between the source voltage and thecurrent in the circuit (Figure 8-5).
Figure 8-5. In an AC circuit, a properly chosen compensating capacitor improves the
power transfer to the load by reducing the phase shift incurred by the impedance of the
transmission line.
Numerical Example
Let us begin with the circuit of Figure 8-2 (neglecting the line impedance) and assume:
( )T rms
T
Load
100 V 0
Z 10
Z 10
=
=
=
V
Thus we obtain the current through the circuit, the voltage across the load resistance, and
the average power transferred to the load as the following:
( ) ( )
( )Load
Load
rmsTrms
T Load
Load rms
Av, Load
100 V 0 5 A 0Z Z 10 10
Z 50 V 0
P * 250 W
= = =
+ +
= =
= =
VI
V I
V I
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Lab 8: AC Power
Now, let us take the line impedance into account as in the circuit of Figure 8-4, andassume that it equals:
( )LineZ 5 j200= +
Thus the current through the circuit, the voltage across the load resistance, and theaverage power transferred to the load can be found as the following:
( )( )
( )
( ) ( )Load
Load
rmsTrms
T Line Load
Load rms
Av, Load
100 V 00.06154 j0.4924 A
Z Z Z 10 5 j200 10
Z 0.6154 j4.924 V
P * 0.6154 j4.924 0.06154 j0.4924 2.462 W
= = =
+ + + + +
= =
= = + =
VI
V I
V I
The average power dropped by a factor of >100.
Now, let us add a compensating capacitor to the circuit, as shown in Figure 8-5, and
ensure that its impedance exactly compensates the imaginary part of the line impedance;
in this example:
( )Compensating
Compensating Line
Z j200
Z Z j200 5 j200 5
=
+ = + + =
Then the total circuit impedance becomes purely resistive; in this example:
( )Total T Line Compensating Load
Total
Z Z Z Z Z
Z 10 5 j200 j200 10 25
= + + +
= + + + =
Therefore, the current through the circuit becomes in phase with the source voltage, and
the average power transferred to the load dramatically increases. The results are:
( )
( )( )
( )Load
Load
rmsT
T Line Compensating Load
rms
rms
Load rms
Av, Load
100 V 0
Z Z Z Z 10 5 j200 j200 10
100 V 04 A 0
25
Z 40 V 0
P * 160 W
= =
+ + + + + +
= =
= =
= =
VI
I
V I
V I
Thus the addition of the properly chosen compensating capacitor leads, in this example,
to a 65-fold increase of the average power transferred to the load.
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Lab 8: AC Power
This example clearly demonstrates the importance of phase shift and its compensation for
the optimization of power transfer to the load.
Note that exact compensation of the imaginary part of the line impedance (+j200 in the
example above) by the capacitors impedance (j200 in the example above) is
achieved at a particular frequency, at which the total circuit impedance is purely resistive[you will soon learn that it is the resonant frequency of the circuit].
At a higher frequency, the absolute value of the inductive impedance L increases, and
the absolute value of the capacitive impedance1
Cdecreases, thus the total circuit
impedance
Total T Line Compensating LoadZ Z Z Z Z= + + +
deviates from the purely resistive and becomes inductive (positive imaginary part). Bythe same token, at a lower frequency, the capacitive impedance dominates and the total
circuit impedance becomes capacitive (negative imaginary part).
As a result, varying the frequency of the sinusoidal source voltage leads, in general, tovariations of the average power delivered to the load.
Sneak Preview of the Lab Work
In the lab, you will study effects of the phase shifts on the power transfer using the
function generator as the source of signals and a small speaker as the load. Your signalfrequencies will be in the audio range thus you will clearly hear when the sound becomes
louder. In other words, you will not only calculate the power values but also experience
the difference between poor and effective power transfer.
For lab experiments, you will connect the function generator to the speaker through an
audio transformer (step-down), which is needed to reduce the effect of loading. The roleof the transformer is briefly explained at the end of this introduction; for extra credot
Explorations, you will be offered to repeat some of the experiments in the circuit withouta transformer.
The small speaker, which you will use as the load for your circuits, has 8
nominalresistance and relatively small impedance. In many calculations, you will consider this
speaker as a purely resistive load.
You will begin lab experiments with a circuit that has a transformer without the load and
measure its output, the open circuit voltage, as shown in Figure 8-6.
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Lab 8: AC Power
Figure 8-6. The circuit with a function generator as the voltage source and an audio
transformer.
In this lab, you will use 3 types of circuit diagrams:(1)In the lab, you will use circuits with an audio transformer, as shown in Figure 8-6(2)In the pre-lab, you will use simplified model circuits that neglect the source
resistance, such as shown in Figure 8-7
Figure 8-7. Simplified model for Circuit #0, which you will use in the pre-lab.
(3)In the post-lab, you will use the Thevenin equivalent circuits such as shown inFigure 8-8.
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Lab 8: AC Power
Figure 8-8. The Thevenin equivalent of Circuit #0, which you will use in the post-lab.
As in all your studies of AC signals, you will use the oscilloscope to measure the
voltages. As soon as you connect the speaker (see Figure 8-9), your oscilloscope will
show a notable decrease of the voltage at the terminals a and b. In other words, thevoltage across the speaker is smaller than the open-circuit voltage:
Speaker Open CircuitV V<
Since your speaker acts nearly as a resistive load, the voltage across it is in-phase with
the current through the circuit. By monitoring the function generators signal in one
channel of your oscilloscope and the voltage in the other channel, you can
determine the time delay (and, therefore, the phase shift) between the source voltage andthe current through the circuit.
SpeakerV
Figure 8-9. The circuit with a function generator and an audio transformer with a speakeras the load.
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Lab 8: AC Power
In the pre-lab, you will simulate the circuit with a speaker as the load by using its
simplified model, which is shown in Figure 8-10.
Figure 8-10. Simplified model for a circuit with a speaker, which you will use in the
pre-lab.
In the post-lab, you will use the Thevenin equivalent of the circuit with a speaker (Figure
8-11); from the resistance of the speaker and the voltage across the speaker V BSPB, which
you measure in the lab, you will calculate the equivalent resistance of the source .S, EQR
Figure 8-11. The Thevenin equivalent circuit with a speaker, which you will use in the
post-lab.
To study the effect of the line impedance, you will add an inductor in series with yourspeaker: see Circuit #2 shown in Figure 8-12.
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Lab 8: AC Power
Figure 8-12. The circuit with an inductor in series with the speaker, as a model of the line
impedance.
As you expect from the explanations and the numerical example above, addition of theinductor incurs a phase shift between the source voltage and the current in the circuit and
greatly reduces the power transferred to the speaker. Since the inductors impedance is
frequency-dependent, these effects also depend on the frequency.
In the pre-lab, you will model Circuit #2 using its simplified model, which is shown in
Figure 8-13.
Figure 8-13. Simplified model of the circuit with a speaker and an inductance, which you
will use in the pre-lab.
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Lab 8: AC Power
For the analysis of lab data obtained on Circuit #2, you will use its Thevenin equivalent,
which is shown in Figure 8-14.
Figure 8-14. The Thevenin equivalent of the circuit with an inductance and a speaker,
which you will use in the post-lab.
The next step is to study the effect of a compensating capacitor, which you also add in
series with the speaker: see Circuit #3 shown in Figure 8-15.
Figure 8-15. The circuit with a compensating capacitor in series with the inductor and the
speaker.
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Lab 8: AC Power
As before, you will perform measurements at various frequencies and observe that, at a
particular frequency, the phase shift between the source voltage and the current in thecircuit vanishes.
In the pre-lab, you will use the simplified model of Circuit #3, which is shown in Figure
8-16.
Figure 8-16. The simplified model of the circuit with a speaker, inductor and a
compensating capacitor, which you will use in the pre-lab.
For the post-lab analysis of your data, you will use the Thevenin equivalent of Circuit #3,which is shown in Figure 8-17. Note that, in order to achieve a better agreement between
the model and your data, we take into account the inductors resistance, which you willalso measure in the lab.
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Lab 8: AC Power
Figure 8-17. The Thevenin equivalent of the circuit, which includes the speaker, inductor,
and a compensating capacitor.
For extra credit Explorations, you will be offered two types of lab experiments. In thefirst Exploration, you will double the capacitor in the circuit as shown in Figure 8-18.
Figure 8-18. The circuit offered for Explorations in the lab.
For further Exploration, you can pull out the transformer and connect your circuit directlyto the function generator as the source: see Figure 8-19.
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Lab 8: AC Power
Figure 8-19. The circuit without a transformer, offered for further Explorations in the lab.
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Lab 8: AC Power
The Role of a Transformer (Optional Reading)
A simple transformer has 2 coils of wire wound on the same magnetic core (Figure 8-20).
Figure 8-20. A transformer with two coils.
If a sinusoidal voltage source is connected to the primary coil of a transformer, itproduces the output voltage at the same frequency in the secondary coil.
The ratio of voltages in its primary and secondary lines is proportional to the ratio of the
numbers of turns in the primary and secondary coils, as shown in Figure 8-21.
An ideal transformer does not absorb power thus the power in the secondary line equals
the power in its primary line; therefore, the ratio of currents is inverse of the ratio of
voltages.
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Lab 8: AC Power
Figure 8-21. Voltages and currents in an ideal transformer.
If the number of turns in the primary coil is smaller than that in the secondary coil, thenthe output voltage amplitude is greater than the input:
1 2
1 2
n n
V V
we call it a step-down transformer. Note that up and down belong to the voltages; the
ratio of currents is inverse. Thus, in a step-down transformer, the current in the secondary
coil is larger than the current in the primary coil.
In many cases, the user decides which coil of the transformer should be primary and
which coil will be secondary. In other words, many transformers can be used either as
step-up or step-down.
If the primary line is connected to a voltage source and the secondary line is
connected to the load, the current in the secondary line is determined by the voltage
and the load impedance (see Figure 8-22). For simplicity, we consider a purely
resistive load.
1V
2V
LoadZ
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Lab 8: AC Power
In the case of a step-down transformer, the current in the primary coil is smaller than the
current in its secondary coil, although the voltage in the primary coil is higher. In otherwords, the apparent load, which is calculated from the ratio of voltage and current in the
primary coil, is larger than the load, which is actually connected to the circuit
(Figure 8-22).
Figure 8-22. The relationship between the apparent load and the actual load.
It means that the source connected to a step-down transformer sees the load as if it had
a much larger resistance.
Loading is the effect of a small load on the output voltage. It is reduced if the load
increases, as in a circuit with a step-down transformer.
Note that the analysis above is simplistic: it neglects the resistance of the transformers
coils.
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Lab 8: AC Power
Pre-Lab:
1.Direct Speaker ConnectionTo model the circuit with a speaker as the load, you will use Circuit #1-MODEL forPart 1.
Assume VBS, EQB = 400 mVBPPKB
Use the speakers nominal resistance: RBSPB = 8
A.Use Multisim to simulate Circuit #1-MODEL. Fill in the following table, wheretBSHIFTB is the time delay between VBSB and VBSPB, and PBAVGB is the average power absorbedby the speaker:
Frequency (Hz) VB
SP, PPKB
(mV) tB
SHIFTB
(s) PB
AVGB
(mW)1500
2500
3500
4500
B.Does PBAVGB depend on the frequency? Why or why not?
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Lab 8: AC Power
2.Effects of InductanceUse Circuit #2-MODEL for Part 2.
Assume VBS, EQB = 400 mVBPPKB
RBSPB = 8
L = 10 mH
A.Use Multisim to simulate Circuit #2-MODEL. Fill in the following table, wheret
B
SHIFTB
is the time delay between VB
SB
and VB
SPB
, and PB
AVGB
is the average power absorbedby the speaker:
Frequency (Hz) VBSP, PPKB (mV) tBSHIFTB (s) PBAVGB (mW)
1500
2500
3500
4500
B.Does PB
AVGB depend on the frequency? Why or why not?
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Lab 8: AC Power
3.Effects of CapacitanceUse Circuit #3-MODEL for Part 3.
Assume VBS, EQB = 400 mVBPPKB
RBSPB = 8 L = 10 mH
C = 0.22 F
A.Use Multisim to simulate Circuit #3-MODEL. Fill in the following table, wheretBSHIFTB is the time delay between VBSB and VBSPB, and PBAVGB is the average powerabsorbed by the speaker:
Frequency (Hz) VBSP, PPKB (mV) tBSHIFTB (s) PBAVGB (mW)
1500
2500
3500
4500
B.Does PBAVGB depend on the frequency? Why or why not?
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Lab 8: AC Power
In-Lab Work
Part 1: Circuit Parameters and Direct Speaker
Connection Obtain a 10 mH inductor, a 8 speaker, and an audio transformer from your instructor.
[We use a Triad MIL-T-27E SP-48 Audio Transformer.]
Turn on the NI ELVIS II.
From the NI ELVISmx Instrument Launcher, launch the DMM.
Use the DMM VI to measure the following components:
RBLB (resistance of the inductor) = ______
RBSP B(resistance of the speaker) = ______
L (inductance of the inductor) = _____ mHLBSPB (inductance of the speaker) = ______ mH
Build the following Circuit #0:
Leave the speaker disconnected for now.
From the NI ELVISmx Instrument Launcher, launch the FGEN and OSCOPE.
On the FGEN VI, create a 1.5 kHz, 10 V BPPKB sine wave with 0 V DC offset.
Power on the PB.
Use the OSCOPE VI to measure VBSB and VBOpen CircuitB. Adjust the OSCOPE parameters so
that you clearly view the waveforms.
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Lab 8: AC Power
Record the peak value of the Open Circuit voltage: V BOCB = _______ VBPPKB
Power off the PB.
Connect the speaker to the circuit at the nodes labeled a and b, to form Circuit #1.
Power on the PB.
Adjust the OSCOPE so that you can clearly view the waveforms.
Use the cursors to measure the phase shift between V BINB and VBSPB.
Record the value of tBSHIFTB and VBPPKB in the following table:
Frequency (Hz) VBSP, PPKB (V) tBSHIFTB (s) Phase Shift ()
1500
2500
3500
4500
Adjust the frequency of the FGEN VI to each of the other frequencies and repeat the
measurements, to complete the table.
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Lab 8: AC Power
Part 2: Effects of Inductance Build the following Circuit #2:
Use L = 10 mH
Create a 1.5 kHz, 10 V BPPKB sine wave with 0 V DC offset on the FGEN.
Adjust the OSCOPE parameters so that you can clearly view the waveforms.
Use the cursors to measure the phase delay between VBSB and VBSPB
Record the VBSP, PPKB and tBSHIFTB in the following table:
Frequency (Hz) VBS, PPKB (V) VBSP, PPKB (V) tBSHIFTB (s) Phase Shift()
1500
2500
3500
4500
Adjust the frequency of the FGEN VI to each of the other frequencies and repeat themeasurements, to complete the table.
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Lab 8: AC Power
Part 3: Effects of Capacitance Build the following Circuit #3:
Use L = 10 mH
C = 0.22 F
Create a 1.5 kHz, 10 V BPPKB sine wave with 0 V DC offset on the FGEN.
Adjust the OSCOPE parameters so that you can clearly view the waveforms.
Use the cursors to measure the phase delay between VBSB and VBSPB
Record the VB
SP, PPKB
and tB
SHIFTB
in the following table:
Frequency (Hz) VBS, PPKB (V) VBSP, PPKB (V) tBSHIFTB (s) Phase Shift()
1500
2500
3500
4500
Adjust the frequency of the FGEN VI to each of the other frequencies and repeat the
measurements, to complete the table.
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Lab 8: AC Power
Part 4: Explorations More Capacitance Build the following Circuit #4:
Use L = 10 mH
C = 0.22 F
Create a 1.5 kHz, 10 V BPPKB sine wave with 0 V DC offset on the FGEN.
Adjust the OSCOPE parameters so that you can clearly view the waveforms.
Use the cursors to measure the phase delay between VBSB and VBSPB
Record the VB
SP, PPKB
and tB
SHIFTB
in the following table:
Frequency (Hz) VBS, PPKB (V) VBSP, PPKB (V) tBSHIFTB (s) Phase Shift()
1500
2500
3500
4500
Adjust the frequency of the FGEN VI to each of the other frequencies and repeat the
measurements, to complete the table.
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Lab 8: AC Power
Part 5: Explorations Without the Transformer Build the following circuit:
Where L = 10 mH
C = 0.22 F
Create a 1.5 kHz, 0.4 V BPPKB sine wave with 0 V DC offset on the FGEN.
Adjust the OSCOPE parameters so that you can clearly view the waveforms.
Use the cursors to measure the phase delay between VBSB and VBSPB
Record the VB
SP, PPKB
and tB
SHIFTB
in the following table:
Frequency (Hz) VBS, PPKB (V) VBSP, PPKB (V) tBSHIFTB (s) Phase Shift()
1500
2500
3500
4500
Adjust the frequency of the FGEN VI to each of the other frequencies and repeat the
measurements, to complete the table.
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Lab 8: AC Power
Post-Lab:
1.Direct Speaker ConnectionIn the lab, you built Circuit #1:
In the pre-lab, you modeled it with Circuit #1-MODEL:
In the post-lab, you will use the Thevenin equivalent Circuit #1-EQ for data analysis:
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Lab 8: AC Power
A.From your lab data, calculate the average power delivered to the speaker for eachfrequency (1500, 2500, 3500, 4500 Hz). Use the actual speaker resistance measured
in the lab for this calculation.
B.Compare your results from Part 1.A above and your in-lab data to your pre-labcalculations (Part 1.A from the Pre-Lab). Specifically, compare the values for PBAVGBand tBSHIFTB. Discuss their agreement/disagreement. Explain possible sources of error.
C.Use your lab data for VBOCB and VBSPB from In-Lab Part 1 to calculate the Theveninequivalent resistance of the transformer output. Repeat your pre-lab simulation usingthe following circuit:
Use VBSB = VBOCB which you measured in In-Lab Part 1
Complete the following table:
Frequency (Hz) VBSP, PPKB (mV) t
BSHIFTB (s) PBAVGB (mW)
1500
2500
3500
4500
D. Compare your results from Part 1.A above and your in-lab data to your newsimulation results (1.C from the Post-Lab). Specifically, compare the values for PBAVGB
and tBSHIFTB. Discuss their agreement/disagreement. Does accounting for the Thevenin
resistance improve the circuit model?
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Lab 8: AC Power
2.Effects of InductanceIn the lab, you built Circuit #2:
In the pre-lab, you used its simplified model:
In the post-lab, you will use the Thevenin equivalent Circuit #2-EQ for data analysis:
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Lab 8: AC Power
A.From your lab data, calculate the average power delivered to the speaker for eachfrequency (1500, 2500, 3500, 4500 Hz). Use the actual speaker resistance measuredin the lab for this calculation.
B.Compare your results from Part 2.A above and your in-lab data to your pre-labcalculations (Part 2.A from the Pre-Lab). Specifically, compare the values for PBAVGB
and tBSHIFTB. Discuss their agreement/disagreement. Explain possible sources of error.C.Repeat the pre-lab simulation using Circuit #2-EQ:
Use VBSB = VBOCB which you measure in In-Lab Part 1.
Complete the table.
Frequency (Hz) VBSP, PPKB (mV) tBSHIFTB (s) PBAVGB (mW)1500
2500
3500
4500
D.Compare your results from Part 2.A above and your in-lab data to your newsimulation results (Part 2.C above). Specifically, compare the values for PBAVGB and
tBSHIFTB. Discuss their agreement/disagreement. Explain possible sources of error.
E. Discuss whether taking RBLB and RBTB into account improved the circuit model. Why orwhy not?
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Lab 8: AC Power
3.Effects of CapacitanceIn the lab, you built Circuit #3:
In the pre-lab, you used its simplified model:
In the post-lab, use the Thevenin equivalent Circuit #3-EQ for data analysis:
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Lab 8: AC Power
A.From your lab data, calculate the average power delivered to the speaker for eachfrequency (1500, 2500, 3500, 4500 Hz). Use the actual speaker resistance measured
in the lab for this calculation.B.Compare your results from Part 3.A above and your in-lab data to your pre-lab
calculations (Part 3.A from the Pre-Lab). Specifically, compare the values for PBAVGB
and tBSHIFTB. Discuss their agreement/disagreement. Explain possible sources of error.C.Repeat the pre-lab simulation with the Thevenin equivalent Circuit #3-EQ.
Use VBSB = VBOCB which you measured in In-Lab Part 1.
Complete the table.
Frequency (Hz) VBSP, PPKB (mV) tBSHIFTB (s) PBAVGB (mW)
1500
2500
3500
4500
D.Compare your results from Part 3.A above and your in-lab data to your newsimulation results (Part 3.C above). Specifically, compare the values for PBAVGB and
tBSHIFT
B. Discuss their agreement/disagreement. Explain possible sources of error.
E. Discuss whether taking RBLB and RBTB into account improved the circuit model. Why orwhy not?
F. Discuss whether adding the capacitor helped transfer more power to the speaker(relative to Circuit #2). Why or why not?
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Lab 8: AC Power
4.Exploration: More CapacitanceIn the lab, you built Circuit #4:
A.From your lab data, calculate the average power delivered to the speaker for eachfrequency (1500, 2500, 3500, 4500 Hz). Use the actual speaker resistance measured
in the lab for this calculation.B.Explain your results.
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Lab 8: AC Power
5.Exploration: Without the TransformerIn the lab, you built the following circuit:
A.From your lab data, calculate the average power delivered to the speaker for eachfrequency (1500, 2500, 3500, 4500 Hz). Use the actual speaker resistance measuredin the lab for this calculation.
B.Explain your results.