ELEC166
Tutorial Week 4 Solutions
Q1.
A:
(a)
(b)
(c)
(d)
Define frequency. What is the unit of frequency?
Define peak amplitude.
What is angular frequency? How is it related to frequency?
What is the rms amplitude of a voltage waveform? How is the rms
amplitude of a sinusoidal voltage waveform related to its peak
amplitude? How about for waveforms which are not sinusoidal?
See section 3.1 of the Lecture Notes.
Q2. On the same graph:
(a) Draw two cycles of a sine wave in black.
(b) Draw in blue (or a dotted line) another sine wave which is +45° out of
phase with respect to the black sine wave.
(c) Draw in red (or a dashed line) another sine wave which is +405° out of
phase with respect to the black sine wave.
(d) What do you notice about the phase difference between the red and the
blue sine waves?
A:
(a), (b) You should have two sine waves, with the blue (or dotted) sine wave
1/8 of a cycle ahead of the black one.
(c), (d) The red and blue waveforms should be identical, since they are 360°
out of phase with each other (415 = 360 + 45). So your graph should look like
this (the actual amplitude values are not important):
Sine waves 2 and 3
(+45°, +405°)
Sine wave 1
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
0
0.5
1
1.5
2
Number of cycles
Q3. Three AC (sinusoidal) signals vary with frequencies of 50 Hz, 100 Hz and 200
Hz. Calculate their periods then, on the same labelled graph with a time
range of 0 to 20 ms, sketch the signal waveforms.
ELEC166 Tutorial Week 4 Solutions Page 1
A:
Your graph should look like the one below (the actual amplitude values are
not important). Note that the waveforms become more “compressed” in time
as the frequency rises.
50Hz
100Hz
200Hz
1
0.5
0
-0.5
-1
0
0.005
0.01
0.015
0.02
Time (seconds)
Q4. A 100pF capacitor is connected in series with a 240V 50Hz AC voltage source.
What is the AC current flowing through the capacitor?
A:
1
, which gives a
ωC
value of 3.18×107Ω. We can use the reactance together with the magnitudes
(amplitudes) of the voltage and current in a way similar to Ohm’s Law, so
that the magnitude of the current is given by the magnitude of the voltage (in
this case 240V rms) divided by the reactance. This gives the magnitude of the
current as 7.5µA. This will be an rms quantity, since the voltage was given as
an rms quantity. Note that this type of calculation does not tell us anything
about the relative phases of any voltages or currents.
The reactance of the capacitor at 50 Hz is given by X C =
Q5. Draw a circuit where a 6V DC voltage source is connected in series with a
switch, a 1kΩ resistor and a 3nF capacitor which is initially uncharged. If we
close the switch at time t = 0, approximately how long will it take for the
voltage across the capacitor to reach 4V? What is the voltage across the
resistor at that time?
A:
The circuit will look like this (the component order is unimportant):
1kΩ
6V
3nF
Eventually, the capacitor will charge up to a voltage which is extremely close
to the battery voltage; in the shorter term it will charge up to about 2/3 of the
ELEC166 Tutorial Week 4 Solutions Page 2
battery voltage in one time constant, a time equal to R×C. Plugging in values
gives the time constant to be 3 µs, so the capacitor will charge up to about 4V
in 3 µs. The voltage across the resistor must be about 2V, since the voltages
across the capacitor and resistor must always add up to 6V.
Q6. What range of capacitance do you need to tune a 100 nH inductor to resonate
over the whole range of the FM broadcast band (88 to 108 MHz)?
A:
1
1
, giving C = 2 . At 88 MHz,
LC
ω L
6
8
ω = 2π×88×10 = 5.53×10 rad/sec, and at 108 MHz, ω = 6.79×108 rad/sec.
Plugging values in gives C = 3.27×10-11 to 2.17×10-11 F, or 32.7 to 21.7 pF.
For a tuned circuit we use the formula ω 2 =
Q7. What is the output voltage of this voltage divider?
3Ω
5 sin(50t)
2Ω
A:
Output
When there are no inductors or capacitors in a circuit, it behaves identically
for AC and DC at each instant of time, so we can use exactly the same
2Ω
equations. The output voltage is then 5 sin(50t ) ×
= 2 sin(50t ) .
3Ω + 2Ω
Q8. An AC voltage waveform varies with time according to the relationship
v(t) = 10 sin(60t+0.5).
(a)
(b)
(c)
(d)
(e)
A:
What are the maximum and minimum values of v(t)?
What is the peak-to-peak amplitude of v(t)?
What is the rms amplitude of v(t)?
What is the frequency of v(t)?
What is the period of v(t)?
(a) +10V, −10V (b) 20V (twice the peak value) (c) 10/√2=7.07V (we can use
this formula since it is sinusoidal) (d) Comparing the equation to the
standard form v(t)=A sin(ωt+φ), we see that ω = 60 rad/sec, and thus f =
60/(2π) = 9.55 Hz (e) T = 1/f = 0.105 s
Q9. When the current flowing through an inductor is suddenly turned off, a large
voltage “spike” appears across the inductor. This is often seen as interference
on a television at the time of turning off an appliance, such as a hair-dryer.
ELEC166 Tutorial Week 4 Solutions Page 3
The voltage generated is described by the formula V = L di/dt (or V = L×rate
of change of current). Assume that an electric motor which has an inductance
of 1mH is drawing 5A of current at the moment it is turned off, and the
current drops to zero in 100µs. What voltage would be generated?
A:
Since the current changes from 5A to zero in 100µs, the average rate of
change of current is 5A/100×10-6s=5×104 amps per second. Thus the voltage
generated is 1×10-3H × 5×104A/s = 50V.
Q10. Australia’s household electricity is nominally 240V rms.
corresponding peak-to-peak voltage.
A:
Calculate the
Since the mains voltage waveform is (pretty close to) sinusoidal, the peak
voltage is √2 times the rms value, or approximately 339.4V. So the peak-topeak value will be twice this, or about 679V.
Q11. If the voltage across a normally operating 60W light bulb is described by
v(t) = 300 sin(2π×103t − 0.33), what is the resistance of the bulb?
A:
The peak voltage is 300V, and since it is sinusoidal, the rms value must be
300/√2 = 212V. Rearranging the equation P=V2/R gives a resistance of about
750Ω. Note that when dealing with AC, V in this equation must be the rms
voltage.
Q12. A black box is known to contain only resistors, capacitors and inductors. If
the input is a sinusoidal voltage waveform, what can you say about the
output voltage waveform?
“Black Box”
A:
Output
Basically, this is a linear system. One important property of a linear system
is that if you put in a signal containing only one frequency (that is, a
sinusoidal waveform), then no new frequencies can be produced by the
system. However, the amplitude of the signal can be changed (and since there
are no amplifiers in the box, the output amplitude cannot be larger than the
input amplitude), and a phase shift can occur.
So the output voltage will be sinusoidal, at the same frequency as the input,
but its amplitude will be smaller, and its phase may be different.
Q13. If there is a voltage of 5.55 V across a 2.2 nF capacitor, how much charge is
stored in the capacitor? Give your answer in scientific notation with 3
significant digits.
A:
Using the basic formula Q = C×V with V=5.55 V and C=2.2×10-9 F gives a
charge of 1.22×10-8 coulomb.
ELEC166 Tutorial Week 4 Solutions Page 4
Q14. Given the two voltage waveforms:
v1(t) = 6 sin(200πt + 0.5)
v2(t) = 6 sin(200πt)
(a)
(b)
(c)
(d)
(e)
A:
Calculate the period and the frequency of each.
Calculate the peak and the peak-to-peak amplitude of each.
Calculate the rms amplitude of each.
On the same graph, plot several cycles of v1(t) and v2(t) vs time.
Comment on the plots.
(a) The frequency is 100Hz and the period is 0.01s for both.
(b) The peak value is 6V and the peak-to-peak value is 12V for both.
(c) The rms amplitudes is 6/√2, or about 4.24V for both.
(d) See graph below.
(e) The waveforms are sinusoidal, with v1 leading v2 by 0.5 radians (≈29°).
8
6sin(200πt+0.5)
6
6sin(200πt)
4
2
Voltage (V)
0
-2
-4
-6
-8
0
0.005
0.01
0.015
0.02
0.025
0.03
Time (sec)
Q15. Given a voltage waveform:
v(t) = 4 + 5 sin(20πt)
(a)
(b)
(c)
(d)
(e)
(f)
(g)
A:
Determine the DC voltage component of v(t).
Determine the peak-to-peak amplitude of v(t).
Determine the rms amplitude of the AC component of v(t).
Determine the period and frequency of v(t).
Plot the DC component of v(t) vs time, from t = 0 to t = 0.1s.
On the same graph, plot the AC component of v(t) vs time.
On the same graph, plot v(t) vs time.
(a) The DC voltage component is the average value, which is 7V (the average
value of the sine part − the AC component − is zero).
(b) The maximum voltage is 4+5=+9V and the minimum voltage is 4−5=−1V.
Thus the peak-to-peak amplitude is +9−(−1)=10V. This will be the same
whether you consider the whole waveform, or just the AC component.
ELEC166 Tutorial Week 4 Solutions Page 5
(c) The rms amplitude of the AC component is 5/√2, or 3.54V.
(d) The frequency is 10Hz (=20π/(2π)), and the period is 0.1s.
(e) This will be a horizontal line at V=+4V volts.
(f) This will be one cycle of a sine wave with peak amplitude 5V.
(g) This will be one cycle of a sine wave with peak amplitude 5V, shifted
upwards (that is, offset) by 4V.
Q16. Sketch the output voltage waveform across the capacitor in the following
circuit. The capacitor is initially uncharged.
Input voltage
source waveform
1MΩ
Input voltage
source
V
Output
1µF
0
1
2
time (sec)
HINT: Think about the charging and discharging of the capacitor.
A:
Basically, we need to think about the voltage source charging (or discharging)
the capacitor via the resistor, and to consider the time constant of the circuit.
The time constant is given by R×C and is equal to 1×106Ω times 1×10-6F,
which is equal to 1 second. Note also that the voltage source, when supplying
a voltage of zero, behaves like a short circuit.
Now at t=0s the capacitor voltage is zero, and stays that way until t=1s, when
the input voltage rises. At this point the capacitor starts charging through
the 1MΩ resistor, so that 1 second later (at t=2s), its voltage has risen to
about 2/3 of the peak input voltage.
At t=2s, the input voltage changes back to zero, and the capacitor starts
discharging back through the resistor. Remember that the voltage source is a
short circuit. The time constants for the charging and discharging are the
same, so that the output voltage (that is, the capacitor voltage) will look
something like this:
Output
voltage
0
1
2
3
time (sec)
ELEC166 Tutorial Week 4 Solutions Page 6