Worksheet for Exploration 31.1: Amplitude, Frequency and
Phase Shift
We characterize the voltage (or current) in AC circuits
in terms of the amplitude, frequency (period) and
phase. The sinusoidal voltage of this function
generator is given by the equation V (t) = V0 sin(ωt - φ)
= V0 sin(2πƒt - φ), where V0 is the amplitude, ƒ is the
frequency (ω = 2πƒ is the angular frequency), and φ is
the phase angle (voltage is given in volts and time
is given in seconds). Restart.
To begin with, keep the resistance of the variable resistor equal to zero. Pick values for the voltage
amplitude (between 0 and 20 V), frequency (between 100 and 2000 Hz) and phase angle (between -2π and
2π).
a.
What does the amplitude on the graph correspond to?
b.
If you increase the amplitude, what do you expect to happen? Try it.
c.
Measure the time between two peaks (or valleys) on the graph. This is the period (T). What does 1/T
equal?
Inputs: Amplitude = _____
frequency = ______
phase shift = ______
Measured value: T = _______ 1/T = ________
d.
What do you need to change to increase the time between two peaks? Try it.
e.
Compare the plots when φ = 0 and when φ = 0.5*π. (You can right-click inside the plot to make a copy.)
f.
What happens when φ = π?
g.
Pick a value of φ other than 0. Measure the time, t (measured from t = 0), it takes for the graph to cross
the horizontal axis with a positive slope (going up). φ should be equal to 2πƒt. So, the phase (or phase
shift) tells you how much the graph is shifted from a straight sin2πƒt curve.
Inputs: Amplitude = _____
frequency = _____
phase shift = ______
Measured: t = _______ (for graph to cross the horizontal axis with a positive slope)
2 π ƒ t = __________
h.
Note that when φ = 0.5*π, the plot is a cosine curve. Why?
Now, change the variable resistor. The plot shows both the voltage across the 1000 Ω resistor (blue) and
the voltage supply (red). Kirchhoff's laws hold for any instant of time in an AC circuit.
i.
Use the techniques you learned for DC circuits to calculate the current in the circuit at several different
points.
Inputs: Amplitude = _____
frequency = ______
Rvar = _______
phase shift = ______
i. Use V = IR at any point in time.
t(sec)
j.
V (volts)
I(amps)
Verify that this circuit is simply a voltage divider.
i. Voltage Divider. Expression for Vout as a function of Vsupply and RA and RB:
RA
+
+
Vout
_
RB
Vsupply
_
k.
What value does the variable resistor need to have for the maximum voltage across the 1000-Ω resistor
to be 1/3 of the value of the source?
i.
Rvar = ____________
ii. Calculations (to determine Rvar):
Worksheet for Exploration 31.2: Reactance
Assume an ideal power supply. The reactance, X, of
a circuit element is the ratio between peak voltage
and current so that V = I X. For a resistor, XR = R.
This exploration shows that for an active load like a
capacitor or inductor, the reactance depends on the
frequency as well (voltage is given in volts, current
is given in amperes or milliamps (note graph
labels), capacitance is given in farads, inductance
is given in henries, and time is given in seconds).
Restart.
a.
For a capacitive load, vary the frequency and observe what happens to the current. How is this result
related to the formula for capacitive reactance? The graph shows the voltage across (red) and current
from the power supply (black) as a function of time. Note that you may need to wait for transient effects
to decay if you change the frequency.
High frequency = _______
Current = _______
Low frequency = ________
Current = _______
i. Formula for capacitive reactance (look in your book if needed)?
ii. Connection between observations of current and capacitive reactance:
b.
Double the capacitance and try it again. What happens? Explain your observations in terms of the
capacitive reactance.
Frequency = ________
C = _________
Current = _________
2C = ________
Current = _________
i. Relation to capacitive reactance?
c.
Repeat (a) and (b) with an inductive load. What happens if you double the inductance? Explain your
observations in terms of the inductive reactance.
High frequency = _______
Current = _______
Low frequency = ________
Current = _______
Frequency = ________
L = _________
Current = _________
2L = ________
Current = _________
i. Formula for inductive reactance (look in your book if needed)?
ii. Connection between observations of current and inductive reactance:
If we take this to the limit of ƒ → 0 (DC circuits), then a capacitor is essentially an open circuit. At high
frequencies, the capacitor is essentially a short circuit (acts like a wire with little or no resistance).
d.
Explain these limits in terms of what a capacitor does (stores charge) and how it works.
e.
At low frequencies, is an inductor essentially an open circuit or a short circuit? What about at high
frequencies?
f.
Explain in terms of what an inductor does (in terms of induced current).
Worksheet for Exploration 31.3: Filters
Since the reactance varies with frequency, we can
use capacitors (or inductors) to filter out different
frequencies. The voltage of the source is red, while
the oscilloscope voltage is the blue plot on the graph
(voltage is given in volts and time is given in
seconds).
a.
At very low frequencies does the capacitor have a high or low a reactance?
b.
Therefore, at low frequencies, will the current through the capacitor be large or small?
Capacitor filter: Try Filter 1.
c.
Will this circuit allow high or low frequency signals to reach the oscilloscope? Explain.
d.
e.
Try it.
Is the amplitude of the voltage measured by the oscilloscope bigger at low or high frequencies?
i. Was your prediction correct or incorrect? If incorrect, how do you need to modify your reasoning?
If it is bigger at higher frequencies, then it "allows" high frequencies through more readily than lower
frequencies and it is called a high-pass filter. If it "allows" low-frequencies through, it is a low pass filter.
Look at the circuit for Filter 2.
f.
Do you think this is a high-pass or low-pass filter? Why?
g.
Try it and determine which kind of filter it is.
Many signals are not simply made up of one single frequency. They are a combination of frequencies and
this is where filters are useful. Try a wave function composed of two different frequency waves with the low
pass filter. Try this wave function with the high pass filter. (Note: you can not change the frequency of this
wave function with the slider bar in this animation.)
h.
What is the difference between the oscilloscope signals in the two cases?
i.
Explain.
Worksheet for Exploration 31.4: Phase Angle and Power
Assume an ideal power supply. The graph shows
the voltage (red) as a function of time across the
source and the current (black) through the circuit
(voltage is given in volts, current is given in
milliamperes, and time is given in seconds).
(Note: the initial current versus time graphs are
not centered about 0 because of the initial state of
the capacitor and inductor.) Restart.
To calculate the power dissipated by an RLC
series circuit you can not simply use IrmsVrms because the power supply current and voltage are not in phase
(unlike with a purely resistive load). This is due to the different phase shifts between voltage and current
associated with the capacitors and inductors (the current through all elements must be the same so the
voltages across each are phase shifted from each other, see Illustrations 31.4 and 31.5). Here the
equations for calculating power are
2
2
P = Irms R = IrmsVrmscosφ = Irms Zcosφ,
where Z is the impedance of the series circuit (Vrms/Irms)and φ is the phase shift between current and voltage
defined as (ω = 2πƒ)
2
2 1/2
Z = (R + (ωL - 1/ωC) )
a.
and
cosφ = R/Z.
Pick a frequency value. Find the impedance from Vrms/Irms .
Frequency = _______
Vrms = ______
Irms = ______
Z = ____________
b.
Compare this value with the calculated value found using the equation above.
R = _____
L = _____
C = ______
ω = 2 π f = ______
Z = (R2 + (ωL - 1/ωC)2)1/2 = ________
c.
Calculate the phase shift.
i. cosφ = R/Z so φ =
d.
Compare this calculated value with the value of phase shift measured directly from the graph. To
measure the phase angle, since one period (1/ƒ) represents a phase shift of 2π, measure the time
difference between the peaks of the voltage and current plots and divide by the period (the time
between the peaks of the voltage or the current) to find the percentage of 2π by which the current is
shifted.
T (period) = ____________
t1 = Time at a maximum of voltage = ________
t2 = Time at nearest maximum of current = ________
φ = phase shift = 2 π (t1-t2)/T = _________
e.
What is the power dissipated?
Worksheet for Exploration 31.5: RL Circuits and Phasors
Assume an ideal power supply. The graph shows the voltage as a
function of time across the source (red), the resistor (blue), and the
inductor (green) (voltage is given in volts and time is given in
seconds). Restart.
To analyze the currents and voltages in this circuit, notice that you
cannot simply use V = I X along with the peak current values. You
must account for the phase differences between the voltages and the
currents. One way to account for the phase differences is to describe
the voltage, current and reactance with phasors. Next to the circuit is
an animation that shows the phasor representation of the circuit
elements (this allows us to show the phase difference) where the
inductor voltage (green) is π/2 ahead of the resistor voltage (blue).
The magnitude of each vector is the peak voltage across the element.
a.
Since there is no phase shift between the resistor current and the voltage across the resistor, what
would a phasor for the current look like in the phasor diagram?
b.
Why is the phasor for the inductor π/2 ahead of the resistor voltage (i.e., the current through the
circuit)? (Hint: Does the inductor current lead or lag the voltage (see Illustration 31.4)?)
c.
How does the length of the inductor's phasor change as you change the frequency?
d.
Why does the length change as a function of frequency?
Notice that the phasors rotate at an angular speed of ω = 2 π f. The projection of any given voltage phasor
on the y axis is the voltage across the circuit element at that time.
e.
Pause the animation and explain how you can tell that the phasor diagram matches the voltages across
the different circuit elements shown in the graph. In other words, verify that the y component of the
vector in the phasor animation matches the value of the voltage shown on the graph.
Frequency = _________
Voltage
Time y-component of vectors
Source inductor resistor source inductor resistor
From the vectors on the phasor diagram, we can develop a connection between the peak (or rms) voltage
and the peak (or rms) current, where V0 = I0Z and the phase difference between the voltage and current is
given by φ. On the phasor diagram, V0 (the source voltage-red) is the vector sum of the two voltage
vectors (resistor-blue, and inductor-green) and φ is the angle between V0 and the current (in the same
direction as the resistor voltage phasor on the diagram). Exploration 31.6 develops the use of phasors for
RLC circuits.
Worksheet for Exploration 31.6: RLC Circuits and Phasors
Assume an ideal power supply. The graph shows the voltage as
a function of time across the source (red), the resistor (blue), the
capacitor (green) and the inductor (yellow), as well as the
current through the circuit (black) (voltage is given in volts,
current is given in milliamperes, angles are given in degrees,
and time is given in seconds). Restart.
From the vectors on the phasor diagram, we can develop a
connection between the peak (or rms) voltage and the peak (or
rms) current, where V0 = I0 Z and the phase difference between
the voltage and current is given by φ. On the phasor diagram, V0
(the source voltage-red) is the vector sum of the three voltage
vectors (resistor-blue, inductor-yellow and capacitor-green)
and φ is the angle between V0 and the resistor phasor (since
resistor current and voltage are in phase). The y components of
the phasor vectors are the voltages across the various circuit
elements. See Illustrations 31.6 and 31.7 as well as Exploration 31.5.
a.
Explain the phase difference between the blue, yellow and green vectors in the phasor animation.
b.
Pick a frequency and pause the animation. Verify that the red vector is the vector sum of the other
three vectors.
c.
Pick a frequency and measure φ on the phasor diagram using the pink protractor.
Frequency = _________
i. Current phasor is in phase with the resistor voltage phasor. Why?
ii. Phase shift (angle between current (resistor voltage) and source voltage)
d.
= __________
Explain how you can tell that the phasor animation matches the voltage and current versus time graph
for the circuit.
e.
Also measure the phase angle on the voltage and current graphs. To measure the phase angle, since
one period (1/ƒ) represents a phase shift of 2π, measure the time difference between the peaks of the
voltage and current plots and divide by the period.
T (period) = ____________
t1 = Time at a maximum of voltage = ________
t2 = Time at nearest maximum of current = ________
φ = phase shift=2π*(t1-t2)/T = ________
f.
Measure Z for this same frequency (Z = V0 / I0). Check your answers by using the equations for
impedance and the phase shift between the voltage and the current, Z = (R2 + (ωL - 1/ωC)2)1/2 and
cosφ = R / Z.
Frequency = ________
Vrms = _______ Irms = ________
Z = V0/I0 = Vrms/Irms = ____________
Comparison with R cos φ = ________
and (R2 + (ωL - 1/ωC)2)1/2 = ________
Worksheet for Exploration 31.7: RLC Circuit
Assume ideal components. The graph shows the
voltage across the source (red) and the current
from the source (black) as functions of time
(voltage is given in volts, current is given in
milliamperes, and time is given in seconds).
Restart.
An RLC circuit is similar to an oscillating spring or
child on a swing. If you push the swing at exactly
the same frequency as the natural frequency of oscillation (the most common way to push a swing), it
quickly goes higher and higher. But if you push (or pull) part way through the swing at times that do not
match the natural rhythm of the swing, the swing will not go as high as quickly and might even swing lower
(in a fairly jerky fashion). When the current is the largest, this is called the resonance.
a.
What is the resonant frequency of this circuit?
i. (Adjust the frequency and determine where the current is the largest.)
Resonant frequency = __________
b.
As you move the frequency of the driving source closer to the natural frequency of oscillation, what
happens to the voltage and current?
i. Natural frequency of the circuit is (1/2π)(1/LC)
c.
1/2
.
Pick a new value for the variable resistor. What is the resonant frequency of this circuit?
R = _________
Resonant frequency = __________
d.
What are the differences in the resonances with different values of R?
e.
1/2
Compare the resonant frequency to (1/2π)(1/LC) . It should be the same.
Worksheet for Exploration 31.8: Damped RLC
Assume ideal components. The graph shows the
voltage across the capacitor (red), the voltage across
the inductor (blue), and the voltage across the
resistor (green) as functions of time (voltage is given
in volts).
Change the switches as you explore the behavior of
this circuit. Restart.
a.
Pick a specific time, measure the voltages on the graph, and verify that Kirchhoff's law holds when the
switch is open and when the switch is closed.
Time Capacitor
voltage
Resistor
voltage
Inductor
voltage
∆V (around the circuit loop
containing circuit elements)
b.
What determines the time between peaks of the voltage when you close the switch?
c.
Change the value of the variable resistor. What happens to the time for the oscillations if the resistor is
large? When the resistor is small? Explain.
R
Time for oscillations to damp out