7LABO Ganago Student Lab8

7/28/2019 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

2010 A. Ganago Introduction Page 1 of 20


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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 ) = +


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 ) = +




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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) =


( )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?

2010 A. Ganago Pre-Lab Page 1 of 3


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Lab 8: AC Power

2.Effects of InductanceUse Circuit #2-MODEL for Part 2.


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.


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:


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.

2010 A. Ganago In-Lab Page 1 of 6


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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




Record the VB

SP, PPKB

and tB

SHIFTB

in the following table:


1500

2500

3500

4500





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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




Record the VB

SP, PPKB

and tB

SHIFTB



1500

2500

3500

4500





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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.



Record the VB

SP, PPKB

and tB

SHIFTB



1500

2500

3500

4500





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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:

2010 A. Ganago Post-Lab Page 1 of 8


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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?



32/37

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:



33/37

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?



34/37

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


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.


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:


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.

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