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J. O'Brien; 11/22/02 Page 1 of 3

D. Day 10/18/06 07/10 HEFW

J. O'Brien; XXXXXXXXXXMontgomery College-Rockville-PH231Page

12/1/20

Page 3 of 3

J. O'Brien; 11/22/02

D. Day 10/18/06 07/10 HEFW

PH262

AC Circuits: Resonance

Lab # ____ Name ______________________________________________ Date _________

Lab Partner(s) Name ___________________________________________________________

I. BACKGROUND INFORMATION

When all three types of components are present in the same series circuit, it becomes possible for the inductive and capacitive reactances to at least partially cancel each other. The above graph for such an RLC circuit shows the behavior of the resistor voltage amplitude, and of the phase between current (VR/R) and emf, as the driving frequency is varied over some appropriate range.

Not at ResonanceAt Resonance

The left hand impedance diagram shows a circuit whose inductive reactance is about three times as large as its capacitive reactance. Since inductive reactance (XL) and capacitive reactance (XC) are always 180° out of phase, they are subtracted to obtain the value of the net reactance, X, which is then combined with R in a Pythagorean sum to give

Inspection shows that reducing the frequency from this point will reduce the inductive reactance while increasing the capacitive ... hence some frequency must exist where the two reactances will be equal in size, leaving the impedance equal to the resistance as shown in the right hand diagram. Since this is the lowest possible value for Z in this particular circuit, it also will have the largest possible I, and this maximum current will be exactly in phase with the driving emf.

This condition, called "resonance", occurs at the frequency such that

which is seen to be identical to the natural frequency for undamped oscillations. The power delivered to the resistor by the generator is maximum when the circuit is at resonance.

III. EXPERIMENTAL PROCEDURE

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1.Connect a series combination of L = 22mH, C = .001µF, and R = 470 ohms across the sine generator in such a way that the input voltage and the resistor voltage can be simultaneously displayed on the oscilloscope.

CH1 connected across function generator

CH2 connected across resistor

Voltage across resistor

Input voltage

2.Go to the tool at the bottom left corner of the simulation. Select the sinusoidal waveform for the input signal. Choose the frequency from the menu. Rotate the flywheel at the bottom right corner of the simulation to locate resonance frequency. Look at VR on the o’scope and watch for a maximum as you vary for the input signal. At resonance frequency the amplitude of VR is almost equal to the input signal. These two signals overlap and can be separated. Hover the arrow on the signal screen. Use the zoom in or zoom out in your mouse to separate these signals as depicted below. A time bar scale is given on the x-axis. Adjust the sinusoidal wave in such a way that the crest of a wave is at the beginning of the time bar scale. Measure the length of the time bar scale carefully with the metric ruler. Find 1mm is equal to how many milliseconds. Measure the distance between adjacent hump. To read the period from the scope and compare with the expected value for this LC pair. Repeat for several other LC combinations until you are confident that you understand how to locate the resonance.

Input voltage

Time period

4.6cm=50us

2.7cm= 29.32us

3.Select the frequency from the tool menu. Rotate the flywheel to change the frequency and collect enough data to plot accurate graphs of VR vs . To plot a graph of vs , measure the length of the time bar scale which is given on x-axis. Adjust the sinusoidal wave in such a way that it starts from one side of the time bar scale. Measure the length of the time bar scale carefully with the metric ruler. Find 1mm is equal to how many milliseconds. Measure the time period and time interval . THINK! before you measure. You are graphing a very non-linear curve and you need to carefully choose the points to measure. Note that at resonance () the phase is 0 and VR is maximum and nearly equal to (the signal generator). Be certain to note the frequency values and VR voltages at which phases ±45° are obtained. You will be asked to show in the question at the end of the lab write-up that, at ±45°, the power delivered to the resistor is 1/2 of that delivered at resonance, and hence these frequencies are called the half-power points.

Select the frequency from tool menu

Rotate the flywheel to change the frequency.

length of time bar scale

4.5cm=50us

1cm =11.11us

4.Finally, exchange your resistor for one that is a factor of ~2 or 3 larger. Locate the resonance. To obtain half power points for this circuit. Choose the input voltage (. Find the root mean square value. . Go to the tool at bottom left corner. Select the frequency from the tool menu. First locate the resonance frequency. Then rotate the flywheel counterclockwise to adjust the frequency until reaches to . At this point you will have . Now rotate the flywheel clockwise until reaches to . At this point you will have . You should note that the Df between the half power points is significantly larger than for the lower resistance circuit. A circuit with a narrow resonance peak (a small Df) is called a high-Q circuit, where "Q" is the common abbreviation for the expression "Quality Factor" (in spite of the obvious danger of confusion with "q" for charge). Specifically, Q is given by

and is thus seen to be related to the resonant frequency and to τ (1/e-life) for damped oscillations. Find the Q’s for the ~500 Ω R and for the R that is ~2 or 3 times larger. Compare the values.

A narrow resonance peak is associated with a long time-constant, ie. with low damping. The Q value is of particular importance in most circuit design applications, eg. a radio tuner that must separate signals from two stations broadcasting at carrier frequencies that are close together.

IV. CONCLUSIONS/QUESTIONS

1.Notice that at resonance the signal voltage, e, is pointed along VR (phase = 0). So PR (power dissipated by R) is VR2 / R = e2 / R, and this represents the maximum power dissipation. Then Pmax = e2 / R. Now suppose that = 45º = π/4. Show that PR = e2 / 2R = ½ e2 / R = ½ Pmax. Thus ±45º phase represents the half-power points on the resonance curve, VR vs. ω.

2.In section 5 above you are given the formula for Q for an RLC circuit.

where and

For a mass-spring system we don’t have R, L, and C, but we do have ω0 and τ where and τ = τe = the 1/e lifetime for the amplitude of the oscillation, either electrical or mechanical. Look back at your results for ω0 and τe for the mass-spring system in Lab #2 and calculate Q for the mass-spring system. You should have found that ω0 ≈ 2π rad/sec and τe ≈ 200 s. How does this value of Q compare to the Q for the RLC circuit?

Q

f

f

R

L

C

=

=

=

=

0

0

0

1

2

D

D

w

w

w

t

2

1

0

t

w

=

=

C

L

R

Q

LC

1

0

=

w

R

L

2

=

t

m

K

o

/

=

w

D. Day 10/18/06 07/10 HEFW

J. O'Brien; XXXXXXXXXXMontgomery College-Rockville-PH231Page

12/1/20

Page 3 of 3

J. O'Brien; 11/22/02

D. Day 10/18/06 07/10 HEFW

PH262

AC Circuits: Resonance

Lab # ____ Name ______________________________________________ Date _________

Lab Partner(s) Name ___________________________________________________________

I. BACKGROUND INFORMATION

When all three types of components are present in the same series circuit, it becomes possible for the inductive and capacitive reactances to at least partially cancel each other. The above graph for such an RLC circuit shows the behavior of the resistor voltage amplitude, and of the phase between current (VR/R) and emf, as the driving frequency is varied over some appropriate range.

Not at ResonanceAt Resonance

The left hand impedance diagram shows a circuit whose inductive reactance is about three times as large as its capacitive reactance. Since inductive reactance (XL) and capacitive reactance (XC) are always 180° out of phase, they are subtracted to obtain the value of the net reactance, X, which is then combined with R in a Pythagorean sum to give

Inspection shows that reducing the frequency from this point will reduce the inductive reactance while increasing the capacitive ... hence some frequency must exist where the two reactances will be equal in size, leaving the impedance equal to the resistance as shown in the right hand diagram. Since this is the lowest possible value for Z in this particular circuit, it also will have the largest possible I, and this maximum current will be exactly in phase with the driving emf.

This condition, called "resonance", occurs at the frequency such that

which is seen to be identical to the natural frequency for undamped oscillations. The power delivered to the resistor by the generator is maximum when the circuit is at resonance.

III. EXPERIMENTAL PROCEDURE

Login to everycircuit.com

1.Connect a series combination of L = 22mH, C = .001µF, and R = 470 ohms across the sine generator in such a way that the input voltage and the resistor voltage can be simultaneously displayed on the oscilloscope.

CH1 connected across function generator

CH2 connected across resistor

Voltage across resistor

Input voltage

2.Go to the tool at the bottom left corner of the simulation. Select the sinusoidal waveform for the input signal. Choose the frequency from the menu. Rotate the flywheel at the bottom right corner of the simulation to locate resonance frequency. Look at VR on the o’scope and watch for a maximum as you vary for the input signal. At resonance frequency the amplitude of VR is almost equal to the input signal. These two signals overlap and can be separated. Hover the arrow on the signal screen. Use the zoom in or zoom out in your mouse to separate these signals as depicted below. A time bar scale is given on the x-axis. Adjust the sinusoidal wave in such a way that the crest of a wave is at the beginning of the time bar scale. Measure the length of the time bar scale carefully with the metric ruler. Find 1mm is equal to how many milliseconds. Measure the distance between adjacent hump. To read the period from the scope and compare with the expected value for this LC pair. Repeat for several other LC combinations until you are confident that you understand how to locate the resonance.

Input voltage

Time period

4.6cm=50us

2.7cm= 29.32us

3.Select the frequency from the tool menu. Rotate the flywheel to change the frequency and collect enough data to plot accurate graphs of VR vs . To plot a graph of vs , measure the length of the time bar scale which is given on x-axis. Adjust the sinusoidal wave in such a way that it starts from one side of the time bar scale. Measure the length of the time bar scale carefully with the metric ruler. Find 1mm is equal to how many milliseconds. Measure the time period and time interval . THINK! before you measure. You are graphing a very non-linear curve and you need to carefully choose the points to measure. Note that at resonance () the phase is 0 and VR is maximum and nearly equal to (the signal generator). Be certain to note the frequency values and VR voltages at which phases ±45° are obtained. You will be asked to show in the question at the end of the lab write-up that, at ±45°, the power delivered to the resistor is 1/2 of that delivered at resonance, and hence these frequencies are called the half-power points.

Select the frequency from tool menu

Rotate the flywheel to change the frequency.

length of time bar scale

4.5cm=50us

1cm =11.11us

4.Finally, exchange your resistor for one that is a factor of ~2 or 3 larger. Locate the resonance. To obtain half power points for this circuit. Choose the input voltage (. Find the root mean square value. . Go to the tool at bottom left corner. Select the frequency from the tool menu. First locate the resonance frequency. Then rotate the flywheel counterclockwise to adjust the frequency until reaches to . At this point you will have . Now rotate the flywheel clockwise until reaches to . At this point you will have . You should note that the Df between the half power points is significantly larger than for the lower resistance circuit. A circuit with a narrow resonance peak (a small Df) is called a high-Q circuit, where "Q" is the common abbreviation for the expression "Quality Factor" (in spite of the obvious danger of confusion with "q" for charge). Specifically, Q is given by

and is thus seen to be related to the resonant frequency and to τ (1/e-life) for damped oscillations. Find the Q’s for the ~500 Ω R and for the R that is ~2 or 3 times larger. Compare the values.

A narrow resonance peak is associated with a long time-constant, ie. with low damping. The Q value is of particular importance in most circuit design applications, eg. a radio tuner that must separate signals from two stations broadcasting at carrier frequencies that are close together.

IV. CONCLUSIONS/QUESTIONS

1.Notice that at resonance the signal voltage, e, is pointed along VR (phase = 0). So PR (power dissipated by R) is VR2 / R = e2 / R, and this represents the maximum power dissipation. Then Pmax = e2 / R. Now suppose that = 45º = π/4. Show that PR = e2 / 2R = ½ e2 / R = ½ Pmax. Thus ±45º phase represents the half-power points on the resonance curve, VR vs. ω.

2.In section 5 above you are given the formula for Q for an RLC circuit.

where and

For a mass-spring system we don’t have R, L, and C, but we do have ω0 and τ where and τ = τe = the 1/e lifetime for the amplitude of the oscillation, either electrical or mechanical. Look back at your results for ω0 and τe for the mass-spring system in Lab #2 and calculate Q for the mass-spring system. You should have found that ω0 ≈ 2π rad/sec and τe ≈ 200 s. How does this value of Q compare to the Q for the RLC circuit?

Q

f

f

R

L

C

=

=

=

=

0

0

0

1

2

D

D

w

w

w

t

2

1

0

t

w

=

=

C

L

R

Q

LC

1

0

=

w

R

L

2

=

t

m

K

o

/

=

w

Answered 2 days AfterMay 02, 2021

1. Voltage amplitude = 1v

Capacitance C=2µf

Resistor R = 1kΩ

Frequency of the source = 46 Hz

According to our measurement;

20 cm corresponds to 50 ms

Therefore, 1 cm...

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