Name:
Lab Partners:
Date:
Pre-Lab 9 Assignment
Capacitors, Inductors & Resonance
(Due at the beginning of lab)
Directions: Read over Lab and then answer the following questions about the procedures.
Question 1 What is the effect of an inductor filter on a multi-frequency AC signal?
Question 2 What does resonance mean? Give an example of resonance in a mechanical system.
Question 3 What do you change when you “tune” a radio to your favourite radio station?
Question 4 The unit of resistance is the ohm, denoted by the greek letter Ω.
a) What is the unit of capacitance?
b) What is the unit of capacitive reactance?
c) What is the unit of inductance?
d) What is the unit of inductive reactance?
PHYS-204: Physics II Laboratory
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Name:
Lab Partners:
Date:
Lab 9
Capacitors, Inductors, and Resonance
Objectives
• To understand the frequency response of capacitors and inductors.
• To understand resonance in circuits driven by AC signals.
Overview
In a previous lab you studied the time-dependent behavior of Alternating Current (AC) signals
that exist all around you. Electronic devices in everyday use such as radio receivers and amplifiers, computers and televisions use AC signals that are manipulated in very precise ways.
Resistors, capacitors, and inductors are used in those manipulations and constitute a very
important part of electronic circuits.
This lab continues the investigation of the role of resistors, capacitors and inductors in
circuits powered by AC signals. In Investigation 1, you will study the role of capacitors and
inductors as AC filters where voltage of certain frequency ranges of AC signals are filtered out
while leaving signals of other frequencies relatively unchanged.
In Investigation 2, you will discover the relationship between peak current and peak voltage
for a series circuit containing a resistor, an inductor, and a capacitor and examine the phase
relationship between the current and the applied voltage in such circuits. You will also study
the situation that maximizes the current in the circuit for a narrow range of frequencies at the
“resonance condition”. The phenomenon of resonance is the central concept underlying the
tuning of a radio or television to a particular frequency.
These labs have been adapted from the Real Time Physics Active Learning Laboratories [?].
The goals, guiding principles and procedures of these labs closely parallel the implementations
found in the work of those authors [?, ?, ?].
Investigation 1:
Introduction to AC Filters
In the previous lab, you explored the relationship between impedance (the AC equivalent of
resistance) and the frequency for a resistor, capacitor, and inductor. These relationships play
a very important role in the design of electronic equipment, in particular audio and video
equipment. You can predict many of the basic characteristics of simple audio circuits based on
what you have learned in previous labs.
In this lab you will create circuits that filter out those AC signals of frequencies that lie
outside the range of interest. In this lab, a filter is a circuit that suppresses the voltage of
certain signal frequencies, leaving other frequency ranges relatively unaffected.
You will need the following materials:
• computer-based laboratory system with Logger Pro software.
• voltage probe.
• multimerter.
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• 10 ohm resistor.
• 100 ohm resistor.
• 7 µF capacitor.
• 8 mH inductor.
• Low impedence signal generator.
Activity 1.1: Capacitors as Filters
In this activity, you will examine how a circuit containing a resistor, a capacitor, and a signal
generator responds to signals of various frequencies. Consider the circuit in Fig. 1 with a
resistor, capacitor, signal generator and voltage probe.
C
+
Vsignal (t)
R
∼
AC input
VP1
−
Figure 1: Capacitive filter circuit: C = 7 µF, R = 100 Ω.
Prediction 1.1 Make a qualitative prediction for the behavior of the peak current through the
resistor, Imax , as the frequency of the signal is increased from zero and sketch it in the left panel
below. [Hint: recall that the relation between the impedance of the capacitor and the frequency
of the signal is given by XC = 2πf1 C .]
Imax (A)
Vmax (V)
Prediction 1.2 Make a qualitative prediction for the behavior of the peak voltage across the
resistor, Vmax , as the frequency of the signal is increased from zero and sketch it in the right
panel below.
100
200
300
400
fsignal (Hz)
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100
200
300
400
500
fsignal (Hz)
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Explain how you arrived at your graphs. Is the graph of the current qualitatively similar or
different from the graph of the voltage? Explain your answer.
Test your predictions.
Step 1: Open the experiment file L9A1-1 (Capacitive Filter).
Comment: In this experiment we use just the voltage probe, even though we want to know
about both the current and the voltage. Remember that in the previous experiment you
found that for a resistor, I = V /R , with the same value of the resistance, R, at all
frequencies. Thus we can use the software to calculate the current by dividing the voltage
measurements by R.
Step 2: Measure the actual resistance of the 100 ohm resistor using the multimeter. On the
computer select Modify Column under the Data menu and choose current. In the
formula that defines the current replace the number 1 in the divisor by your measured
value for R.
Step 3: Zero the voltage probe with it disconnected from the circuit.
Step 4: Connect the resistor, capacitor, signal generator and probe as shown in Fig. 1.
Step 5: Set the signal generator to a frequency of 500Hz. Adjust the amplitude control to half
of its maximum value (the mark on the amplitude knob should be vertical).
Step 6: Collect data. When you have a good graph, click on stop to capture the graph.
Step 7: Use the statistics command in the analysis menu to determine the peak voltage and
peak current, and enter these data in table 1.
Fsignal (Hz)
500
450
400
350
300
250
200
150
100
Vmax (V)
Imax (A)
Table 1: Peak voltage and peak voltage across a resistor in a RC circuit subject to a AC signal.
Step 8: Reduce the frequency of the signal generator by 50 Hz. Begin graphing. Again use
the statistics command to determine the peak voltage and c urrent and enter in table 1.
Step 9: Continue making measurements of peak voltage and current at different frequencies,
lowering by 50Hz until reaching 100Hz. If you find a range of frequencies in which small
changes in frequency produce large changes in the peak voltage, you should be sure to
make measurements at several frequencies within that range.
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Imax (A)
Vmax (V)
Step 10: Plot the data from Table 1 on the axes below. Mark scales on the vertical axes.
100
200
300
400
fsignal (Hz)
500
100
200
300
400
500
fsignal (Hz)
Question 1.1 If you could continue taking data up to very high frequencies, what would
happen to the peak current, Imax , through the resistor and the peak voltage, Vmax , across
the resistor?
Question 1.2 At very high frequencies, does the capacitor act pretty much like an open
circuit (an incomplete or broken circuit) or rather like a short circuit (a wire with a very
small or negligible resistance)? Justify your answer.
Question 1.3 What signal frequency does a DC signal have?
Prediction 1.3 What would be the current through and voltage across the resistor if you
replaced the signal produced by the AC signal generator with a DC signal?
Test your prediction by acquiring a DC data point (by using batteries).
Step 11: Replace the signal generator with two D-cells in series.
Step 12: Begin graphing.
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Step 13: Determine the peak voltage and current.
Step 14: Enter this data point in Table 1 and plot it on your graph.
Question 1.4 At very low frequencies, does the capacitor act pretty much like an open circuit
(an incomplete or broken circuit) or rather like a short circuit (a wire with a very small or
negligible resistance)? Justify your answer.
Comment: In a previous lab you learned that a capacitor’s impedance, its capacitive reactance, decreases as the frequency of the AC signal increases and that the impedance of a resistor
is independent of the signal frequency. Note that even though these elements are in series, their
impedances do not simply
p add together since the current and voltage are not in phase. The
actual relation is: Z = R2 + XC2 . Nevertheless, as you can see from this expression for Z,
the impedance of the series circuit decreases as the reactance of the capacitor decreases. Since
XC = 2πf1 C , the impedance of the circuit decreases as the signal frequency increases.
The peak voltage applied to the circuit by the signal generator and peak current are related
to each other by: Vapplied = Imax Z.
The peak voltage applied by the signal generator remains unchanged in the circuit shown
in Fig. 1, and therefore the peak current in the circuit must in crease as the total impedance
decreases. As a result, the peak voltage across the resistor increases as the frequency of the
signal increases. This type of circuit is an example of a ”high-pass” filter.
Activity 1.2: Inductors as Filters
The activity that follows is similar to the previous one except that you will replace the capacitor
with an inductor and determine the filtering properties of the inductor circuit.
Consider the circuit containing a resistor, inductor, signal generator and probes shown in
Fig. 2 below.
L
+
Vsignal (t)
∼
AC input
R
VP1
−
Figure 2: Inductive filter circuit.L = 8mH, R = 10 Ω
Prediction 1.4 Make a qualitative prediction for the behavior of the peak current through the
resistor, Imax , as the frequency of the signal is increased from zero and sketch it in the left panel
below. [Hint: recall that the inductor’s impedance is related to the frequency of the signal by the
expression XL = 2πf L.]
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Imax (A)
Vmax (V)
Prediction 1.5 Make a qualitative prediction for the behavior of the peak voltage across the
resistor, Vmax , as the frequency of the signal is increased from zero and sketch it in the right
panel below.
0
100
200
300
400
fsignal (Hz)
500
0
100
200
300
400
500
fsignal (Hz)
Explain how you arrived at your graphs. Is the graph of the current qualitatively similar or
different from the graph of the voltage? Explain y our answer.
Test your predictions:
Step 1: Open the experiment file L9A1-2 (Inductive Filter).
Step 2: Measure the actual resistance of the 10 ohm resistor using the multimeter. On the
computer select Modify Column under the Data menu and choose current. In the formula
that defines the current replace the number 1 in the formula by your measured value for
R.
Step 3: Zero the voltage probe with the probe disconnected from the circuit.
Step 4: Connect the resistor, inductor, signal generator and probe as shown in Fig. 1.
Step 5: Set the signal generator to a frequency of 100Hz. Adjust the amplitude control to half
of its maximum value (the mark on the amplitude knob should be near vertical).
Step 6: Collect data. When you have a good graph, click on stop to capture the graph.
Step 7: Use the statistics command in the analysis menu to determine the peak voltage and
peak current, and enter the data in table 2.
Step 8: Increase the frequency of the signal generator to 150Hz. Begin graphing. Again use
the statistics command to determine the peak voltage and current and enter in Table 2.
Step 9: Continuing making measurements of peak voltage and current at different frequencies,
as in step 8. Increase each time the frequency by 50Hz until you reach 500 Hz. If you
find a range of frequencies in which small changes in frequency produce large changes in
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Fsignal (Hz)
100
150
200
250
300
350
400
450
500
1000
Vmax (V)
Imax (A)
Table 2: Peak voltage and peak voltage across a resistor in a LR circuit subject to a AC signal.
the peak voltage, you should be sure to make measurements at several frequencies within
that range.
Imax (A)
Vmax (V)
Step 10: Plot the data from Table 2 on the axes below. Mark scales on the vertical axes.
0
100
200
300
400
fsignal (Hz)
500
0
100
200
300
400
500
fsignal (Hz)
Question 1.5 If you continued taking data up to very high frequencies, what would happen to
the peak voltage Vmax , across the resistor, and peak current, Imax , through the resistor?
Question 1.6 At very high frequencies, does the inductor act more like an open circuit (a break
in the circuit’s wiring), or more like a short circuit (a connection with very little resistance)?
Justify your answer.
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Comment: in the last lab you learned that an inductor’s impedance, its inductive reactance,
increases as the frequency of the AC signal increases and that the impedance of a resistor is
independent of the signal frequency. As with the capacitor and resistor, the impedances of the
inductor and resistor do not simply add together.
(Again because the current and voltage are
p
not in phase.) The actual relation is: Z = R2 + XL2 . As you can see from this expression
for Z, the impedance of the series circuit increases as the reactance of the inductor increases.
Since XL = 2πf L, the impedance of the circuit increases as the signal frequency increases.
Again, the voltage applied by the signal generator and the peak current are related to each
other by Vapplied = Imax Z.
In the circuit in Fig. 2, since the peak voltage from the signal generator remains unchanged,
the peak current in the circuit must decrease as the total impedance increases. Therefore, the
peak voltage across the resistor decreases as the frequency of the signal increases. This type of
circuit is an example of a ”low- pass” circuit or filter.
Activity 1.3: Introduction to Audio Speaker Design
Music is sound waves composed of many different frequencies (pitches) superimposed on each
other all at once, and constantly changing. Low notes correspond to sound waves with a low
frequency. High notes correspond to sound waves with a high frequency. A microphone is a
device that can detect sound waves over a wide range of audible frequencies and convert them
to AC signals. Thus, the signal from a microphone generally consists of many AC frequencies
superimposed on each other all at once. (These signals can then be encoded and recorded on a
storage device, such as a CD.)
Some of the ideas you have studied in this investigation are readily applied to the design
of audio speakers. Many audio speakers have at least two separate ”cones” from which sound
emanates. Generally, the small cone, or tweeter, generates the high frequency sound waves,
and the large cone, o r woofer, generates the low frequency sound waves. To best utilize the
properties of the different sized cones, speaker designers use capacitative and inductive filter
circuits to send high frequency signals to the small cone and low frequency signals to the large
cone.
Imagine that you replace the signal generator in the circuits in Fig. 1 or Fig. 2 with a
collection of different frequency signals coming from your CD player. You also have a woofer
and a tweeter. You can connect either in place of the resistor in one of the circuits.
Question 1.7 Into which circuit should you wire the woofer-the capacitive filter circuit or the
inductive filter circuit? (Note: assume that the bigger the voltage across the resistor, the louder
the sound emitted by the speaker.) Briefly explain your reasoning.
Question 1.8 Into which circuit should you wire the tweeter-the capacitive filter circuit or the
inductive filter circuit? Briefly explain your reasoning.
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Investigation 2:
The Series RLC Resonant Circuit
In this investigation you will use your knowledge of the behavior of resistors, capacitors and
inductors in circuits driven by AC signal of various frequencies to predict and then observe the
behavior of a circuit with a resistor, capacitor, and inductor connected in series.
The RLC series circuit you will study in this investigation exhibits a ”resonance” behavior
that is useful for many familiar applications. One of the most familiar uses of such a circuit is
as a tuner used in a radio receiver to tune to a particular radio station.
You will need the following materials:
• computer-based laboratory system with Logger Pro software.
• voltage probe.
• multimeter.
• 10 ohm resistor.
• 7 µF capacitor.
• 8 mH inductor.
• Low impedance signal generator.
Consider the series RLC circuit shown in Fig. 3.
L
+
Vsignal (t)
∼
AC input
R
VP1
−
C
Figure 3: RLC Series Circuit
Prediction 2.1 At very low signal frequencies (near 0 Hz), will the maximum values of I and
V across the resistor be relatively large, intermediate or small? Explain your reasoning.
Prediction 2.2 At very high signal frequencies (well above 100Hz), will the maximum values
of I and V be relatively large, intermediate or small? Explain your reasoning.
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Prediction 2.3 Based on your Predictions 2.1 and 2-2, is there some intermediate frequency
where I and V will reach maximum or minimum values? Do you think they will be maximum
or minimum?
X (Ω)
Prediction 2.4 On the axes below, draw qualitative graphs of XC vs. frequency and XL vs.
frequency. Clearly label each curve. (Hint: base your answers on your observations in the
previous lab and in investigation 1 of this lab.)
f (Hz)
Comment: As we noted earlier in this lab, the relationship between the total impedance,
Z, for a series combination of a resistor, capacitor, and inductor is not just the sum of the
impedances of the three circuit elements. Instead, because of phase differences, Z is given by
the following expression:
p
Z = R2 + (XL − XC )2
Note that the combination XL − XC appears because the phase current is behind the phase
of the voltage in the inductor, while the phase of the current is ahead in the capacitor.
Prediction 2.5 For what values of XL and XC will the total impedance of the circuit, Z, be
a minimum? On the axes above, mark and label the frequency where this occurs. Explain your
reasoning.
Prediction 2.6 At the frequency you labeled, will the value of the peak current, Imax , in the
circuit be a maximum or minimum? What about the value of the peak voltage, Vmax , across the
resistor? Explain your reasoning.
The point you identified for Predictions 2.5 and 2.6 is the resonant frequency. Label it with
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the symbol f0 . The resonant frequency is the frequency at which the impedance of the series
combination of a resistor, capacitor and inductor is minimum. This occurs at a frequency where
the values of XL and XC are equal.
Prediction 2.7 On the axes above draw a curve that represents XL − XC vs. frequency. Be
sure to label it.
Prediction 2.8 Use your results from Predictions 2.5 and 2.7 to determine the general expression for the resonant frequency, f0 , as a function of L and C. (Hint: you will need the
expressions for XL and XC given below.)
XL = 2πf L
and
XC = 1/(2πf C)
Test your predictions:
Activity 2.1: The Resonant Frequency of a Series RLC circuit
Step 1: Open the experiment file L9A2-1 (RLC Resonance).
Step 2: On the computer select Modify Column under the Data menu and choose current.
In the formula that defines the current replace the number 1 in the formula by the value
you measured for the resistance of your 10 Ω resistor.
Step 3: Zero the voltage probe while disconnected from the circuit.
Step 4: Connect the circuit with resistor, capacitor, inductor, signal generator and probes
shown in Fig. 3.
Step 5: Set the signal generator to a frequency of 100Hz. Adjust the amplitude control to half
of its maximum value (the mark on the amplitude knob should be near vertical).
Step 6: Collect data. When you have a good graph, click on stop to capture the graph.
Step 7: Use the statistics command in the analysis menu to determine the peak voltage and
peak current, and enter these in Table 3.
Step 8: Increase the frequency of the signal generator to a value greater than 100 Hz. Begin
graphing. Again use the statistics command to determine the peak voltage and current
and enter them in Table 3.
Step 9: Continuing making measurements of peak voltage and current at different frequencies,
as in step 8. You should have measurements over the range of frequencies from 100 Hz up
to 1000 Hz. If you find a range of frequencies in which small changes in frequency produce
large increases or decreases in the peak voltage, you should be sure to make measurements
at several frequencies within that range.
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Fsignal (Hz)
Vmax (V)
Imax (A)
Table 3: Peak voltage and peak voltage across a resistor in a LRC circuit subject to a AC
signal.
Imax (A)
Vmax (V)
Step 10: Plot the data from Table 3 on the axes below. Mark scales on the vertical axes.
0
200
400
600
800
fsignal (Hz)
1000
0
200
400
600
800
1000
fsignal (Hz)
Question 2.1 Does the behavior of the voltage across the resistor and current in the
circuit in Fig. 3 agree with your predictions, especially Predictions 2.5 to 2.7? Explain.
Prediction 2.9 Calculate the resonant frequency for your circuit. Show your calculations. (Hint: use the formula from Prediction 2.8 and the actual values of the capacitance
and inductance.)
Hz.
f0 =
Step 11: Measure the resonant frequency of the circuit to within 1 Hz. To do this, begin
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graphing and slowly adjust the frequency of the signal generator until the peak voltage
across the resistor is maximum. (Use your results from Table 3 to help you locate the
resonant frequency.)
f0 =
Hz.
Question 2.2 How does this experimental value for the resonant frequency compare with your
Prediction 2.9?
In a radio receiver, the signal generator is replaced by a long antenna, which picks up all
of the radio signal frequencies. (This is similar to a microphone, mentioned in the previous
activity, detecting all the frequencies, or pitches, of music at once.) By strategically choosing
values of C and L you can tune the circuit to the frequency of your favorite radio station,
meaning the circuit is at resonance (the amplitude of the voltage across R is a maximum) for
that particular station’s frequency. (In many real radio receivers, a variable capacitor is used.
When you turn the knob, you change the capacitance and, hence, the resonant frequency.)
Of course the signal picked up by the radio receiver is very weak-possibly just a microvolt.
An important and surprising property of resonant circuits makes it possible to obtain a much
higher voltage. We will explore this property in the next activity.
Activity 2.2: Voltage across the capacitor at resonance
Question 2.3 Suppose that the peak value of the voltage of the signal applied to the series RLC
resonant circuit is denoted
p by V0 and the total resistance of the circuit is R. You know that
Imax = V0 /Z and Z = R2 + (XL − XC )2 . Using the fact that for the resonant frequency
XL = XC , write an equation relating the peak current at resonance,I0 , to the signal voltage,
V0 , and the total resistance of the circuit, R.
Question 2.4 Does your answer to question 2.3 imply that the current at the resonant frequency depends in any way on the inductance L or the capacitance C in the circuit? Explain
your answer.
Step 1: Remove the connections from the coil in your circuit. Use the multimeter set to
measure ohms to measure the resistance of the coil. Add it to the value you measured for
the resistance to get the total resistance of the series circuit.
Rtot =
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Question 2.5 If the peak value of the signal voltage is V0 , calculate the peak current at
the resonant frequency, using your equation from question 2.4 and the total resistance you
determined in step 1.
Calculated current at resonance, I0 =
A.
Step 2: Reconnect the coil to your resonant circuit.
Step 3: Move the clips of the voltage probe to measure the voltage provided by the signal
generator, as shown in Fig. 4.
L
+
Vsignal (t)
∼
AC input
VP1
R
−
C
Figure 4: RLC resonant circuit with voltage probe connected to measure the applied signal.
Step 4: Set the signal generator to the resonant frequency that you measured in the previous
activity.
Step 5: Collect data. Adjust the amplitude of the signal generator to obtain a peak value of
1 volt for the applied signal. You can be more accurate if you change the scale of the
voltage graph. Stop collecting data after you have the amplitude set.
Warning: Once you have set the amplitude be careful not to move the amplitude control
during the remaining steps in this activity.
Step 6: Move the voltage probe clips to measure the voltage across the resistor, as in Fig. 3.
Step 7: Collect data. When you have a good graph, click on stop to capture the graph.
Step 8: Use the statistics command in the analysis menu to determine the peak current
measured current at resonance: I0 =
A
Question 2.6 Does the measured value of the current at the resonant frequency agree
with the value you calculated in question 2.5?
Question 2.7 The peak AC current flowing through the capacitor and the peak voltage
across the capacitor are related by Ohm’s law, using the impedance of the capacitor: Calculate the voltage that should appear across the capacitor for the current you measured in
the previous step.
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Calculated capacitor voltage at resonance: VC =
V.
Step 9: Move the clips of the voltage probe to measure the voltage across the capacitor, as in
Fig. 5.
L
VP1
Vsignal (t)
R
∼
AC input
−
+
C
Figure 5: RLC resonant circuit with voltage probe connected to measure the voltage across the
capacitor.
Step 10: Collect data. When you have a good graph, click on stop to capture the graph.
Step 11: Use the statistics command in the analysis menu to determine the peak current.
measured capacitor voltage at resonance: VC =
V.
Question 2.8 Does the measured value for the voltage across the capacitor agree with the value
you calculated for question 2.7?
Question 2.9 Can you explain how it is possible for the voltage across just the capacitor to
be larger than the voltage across the entire combination of resistor, inductor and capacitor. In
your explanation you will need to make use of the fact that the phase of the voltage is ahead of
the current in the inductor, and behind in the capacitor. The total phase difference between the
voltage in the inductor and capacitor is almost 180◦ . It may help to draw sketches of two sine
functions out of phase with each other by 180◦ .
This laboratory exercise has been adapted from the references below.
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