Physics 106 - How Things Work - Spring, 1998
Problem Set 2 - Answers
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Due Friday, March 27, 1998, in class
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Please answer each problem as concisely as possible
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You may discuss the problems with one another but you must write them
up separately and in your own words.
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There are 5 problems containing a total of 25 parts. Each part will be
worth 4 points.
Problem 1: Chapter 12, Case 4a-e (Pg. 460)
An uninterruptable power supply (UPS) allows a computer system to operate
during a brief power outage. Most personal computer UPS systems use an
electronic device to convert 12 V DC power from a battery into 120 V AC
power for the computer. The UPS circuitry first converts the 12 V DC into
12 V AC and then uses a transformer to increase the voltage to 120 V, as
required by the computer. In Europe, the final voltage is 220 V.
a. Why can’t the transformer directly convert 12 V DC into 120
V DC?
Answer: Direct current produces only a constant magnetic field
around the transformer's primary coil. Without a changing magnetic field
around that coil, there will be no electric field to push current through
the transformer's secondary coil.
Why: A steady current through the transformer's primary coil
will make that coil magnetic, but its magnetism will be constant. Since
only a changing or moving magnetic field creates an electric field, there
will be no electric field present in the transformer to accelerate charges
in the transformer's secondary coil. Thus the transformer can't transfer
power from a direct current in one circuit to a direct current in another
circuit and can't convert 12 V DC into 120 V DC.
b. An electronic switching system converts the 12 V DC from the
batteries into 12 V AC for the transformer. What is different about the
current flowing through the two wires attached to the battery and the current
flowing through the two wires attached to the transformer?
Answer: The current from the battery always flows in one direction--from
the battery through one wire and back to the battery through the other
wire--while current from the transformer reverses directions frequently.
Why: Alternating current changes directions or "alternates".
The current flows out the first wire and back through the second wire and
then, moments later, it flows out the second wire and back through the
first wire.
c. How does power move from the 12 V AC side of the transformer
to the 120 V AC side of the transformer?
Answer: Current in the 12 V AC side of the transformer produces
a changing magnetic field, which creates an electric field, which pushes
on current in the 120 V AC side of the transformer.
Why: While charge doesn't move from one side of the transformer
to the other, the energy that charges carry does move. It is passed from
the current in the 12 V AC side to the current in the 120 V AC side by
way of a magnetic field and an electric field.
d. Energy is conserved. If there is an average current of about
20 A in the 12 V primary circuit of the transformer, what is the average
current in the 120 V secondary circuit of that transformer?
Answer: 2 A.
Why: When a current of 20 A is flowing through the primary circuit
and experiencing a voltage drop of 12 V as it flows through the transformer
primary coil, it is delivering 20 A times 12 V of power, or 240 W of power
to the transformer. That same 240 W of power should appear in the secondary
circuit, where a current of 2 A experiences a voltage rise of 120 V as
it flows through the transformer secondary coil.
e. The length of time that the UPS can deliver power to the computer
is limited by the battery’s chemical potential energy. Suppose you wish
to extend that time by adding a second 12 V battery to the system. How
should you connect the two batteries together so that the current arriving
at the pair of batteries will still experience a 12 V rise in voltage and
yet be able to extract power from both batteries?
Answer: Attach the two batteries in parallel:
Why: The current entering this pair of batteries will divide
into two portions. Each portion will pass through one battery and emerge
with 12 V more voltage (12 J more energy per coulomb of charge). The two
portions will then join back together into a single current with 12 V more
voltage than before.
Problem 2: Chapter 13, Case 3a-d (Pg. 488)
The power supply in your stereo amplifier uses alternating current from
the electric company to provide direct current to the amplifier. This supply
converts electric power from one form to another through the use of transformers,
diodes, capacitors, and transistors.
a. The power supply provides a relatively small voltage rise
to the large current that it sends through the amplifier. The voltages
provided by the electric company are too high for the amplifier and the
currents are too small. How does a transformer make it possible for the
power supply to provide a relatively small voltage rise to a relatively
large current that flows between it and the amplifier?
Answer: The transformer is a step-down transformer, with relatively
few turns in its secondary coil. The current in the secondary coil makes
relatively few trips around the transformer core and picks up relatively
little voltage. However, a large current can flow through the secondary
coil to pick up this small voltage rise.
Why: The step-down transformer is transfering power from the
primary circuit to the secondary circuit. In the primary circuit, the relatively
few charges that pass through each second are each deliverying large amounts
of energy. In the secondary circuit, the relatively large number of charges
that pass through each second are each picking up small amounts of energy.
Overall, energy is conserved and the power entering the transformer is
equal to the power leaving the transformer.
b. The power supply’s transformer provides AC electric power through
two wires, but the amplifier needs DC electric power through two wires.
To fix this mismatch, the stereo connects these two pairs of wires with
4 diodes so that even though the currents through the two wires of the
transformer reverse, the currents through the two wires of the amplifier
don’t reverse. How are those 4 diodes connected between the transformer
and amplifier? (Draw a picture and indicate which way current can flow
through each diode.)
Answer: Either of these two equivalent drawings:
or
Why: At any given moment, there will probably be positive charge
on one of the AC wires and negative charge on the other. However, you can't
predict which wire will be positive and which will be negative. The 4 diodes
shown above ensure that positive charges can always flow toward the + DC
wire and negative charges can always flow toward the - DC wire. You can
also think of the 2 diodes connected to the - DC wire as allowing current
and positive charges to return to the power company through whichever AC
wire is appropriate at the moment.
c. While the diodes (see b) ensure that current always flows in
one direction through the amplifier, the transformer can’t provide current
when the current from the power company is reversing. To maintain a steady
current through the amplifier, the power supply uses a large capacitor.
When the transformer’s current is large, some of that current is used to
transfer charge between the capacitor’s plates so that it stores separated
charge. When the transformer’s current is small, this separated charge
is allowed to flow through the amplifier as current. Draw a graph of the
voltage difference between the two plates of the capacitor versus time.
Mark the times when the amount of separated charge is increasing and when
it’s decreasing.
Answer:
Why: While the voltage on the AC wires varies as a sine wave,
with the voltage on one wire going first positive then negative with respect
to the other wire, the voltage on the DC wires following the 4-diode "bridge"
always has one wire positive with respect to the other wire. Without the
capacitor, the voltage of the positive wire still varies up and down, depending
on how much charge is available from the AC wires at that moment, but it
never drops below zero. It's voltage is shown above as the dashed line.
But once the capacitor is inserted between the two DC wires and it begins
to store separated charges, the voltage of the positive wire stops returning
all the way to zero during current reversals of the AC wires. Instead,
the voltage of the positive wire sags slowly downward during a current
reversal until enough charge is available again on the AC wires to replenish
the store of separated charges in the capacitor. The voltage on the capacitor
fluctuates up and down somewhat with each half cycle of the AC power line,
rising when more separated charges are available and can flow through the
diodes and falling when there are too few separated charges on the AC wires
to push more separated charges through the diodes to the capacitor.
d. The transformer, diodes, and capacitor do a pretty good job of
sending direct current through the amplifier, but there still tend to be
periodic fluctuations in that current. To keep the current stable, the
amplifier uses a regulating device. Just before the current enters the
amplifier, it passes through an
n-channel MOSFET. An electronic sensor measures the current through
the amplifier and determines if that current is too high or too low. It
then adjusts the charge on the gate of the MOSFET to increase or decrease
the MOSFET’s electric resistance and lower or raise the current through
the amplifier. If the sensor detects that the current is too high, should
it increase or decrease the positive charge on the MOSFET’s gate? Explain.
Answer: The sensor should decrease the positive charge on the
MOSFET's gate, reducing the MOSFET's ability to carry current by allowing
the p-n junction to start reappearing (or, equivalently, making the channel
less n-type and more p-type).
Why: When there is lots of positive charge on the gate of the
n-channel MOSFET, its central region--the channel--behaves as n-type semiconductor.
Since the whole MOSFET is then effectively n-type, current can flow through
it and it's a conductor. When there is no positive charge on the gate,
the MOSFET's channel is p-type semiconductor. Since the MOSFET then contains
two p-n junctions, turned in opposite directions, no current flows through
the device and it's an insulator. But when there is a modest amount of
positive charge on the gate, the MOSFET is somewhere in between a conductor
and an insulator. It then resists the flow of current but doesn't block
that current completely. To decrease the current through the MOSFET, you
want it to resist the current more, so you must decrease the amount of
positive charge on its gate.
Problem 3: Chapter 13, Case 4a-e (Pg. 488)
You enjoy listening to your little portable radio while jogging, but
it hasn’t been working properly since it got wet in the rain last week.
The problem is that it keeps turning itself off. Because the radio makes
no “click” sound when you press the “on” or “off” buttons and because it
turns itself off automatically after an hour, you know that the switch
that controls the radio’s power is electronic. It’s probably an n-channel
MOSFET that’s connected in series with the radio’s electronics so that
current from the battery must pass through both the electronics and the
MOSFET before returning to the battery.
a. Why won’t any power reach the electronics when the MOSFET
isn’t conducting current?
Answer: The circuit will be open and no electric current will
flow through the electronics. With no current flowing, no power can move
from the battery to the radio.
Why: To maintain a flow of power between the battery and the
radio, charges must be able to flow out of the battery with extra energy
through one wire and return to the battery with little energy through a
second wire. Since the MOSFET is part of one of these wires and it isn't
allowing current to flow, this circuit isn't active. Charges either can't
flow out to the radio or they can't return to the battery to pick up more
energy.
b. What must the “on” button do to make the n-channel MOSFET conduct
current so that the radio will operate?
Answer: It must put positive charge on the gate of the MOSFET.
Why: An n-channel MOSFET becomes a conductor when its central
part or "channel" shifts from being p-type to being n-type. This shift
can be caused by attracting negatively charged electrons into the channel
by putting positive charge on the nearby gate.
c. What must the “off” button or the automatic shutdown do to stop
the n-channel MOSFET from conducting current, so that the radio will turn
off?
Answer: It must remove the positive charge from the gate of
the MOSFET (or, equivalently, put negative charge on that gate).
Why: When there is no positive charge on the n-channel MOSFET's
gate, it develops two back-to-back p-n junctions and acts like a pair of
diodes arranged back-to-back. No current can flow through such a structure.
d. Water is a poor conductor of electricity, but with patience you
can send charge through it. If water is slowly turning off the radio, what
is it probably doing?
Answer: The water is allowing positive charge to flow off the
gate (most probably to the negative terminal of the battery).
Why: The water can't store charge, so it is simply acting as
a conduit for that charge. Positive charge is leaving the gate through
the water and going somewhere else. The most logical destination for this
positive charge is the negative terminal of the battery.
e. You discover a small drop of water inside the “off” button, allowing
positive charge to flow slowly from the gate of the n-channel MOSFET to
the negative terminal of the battery. You remove the water and the radio
works perfectly. Why did the drop cause trouble and why did removing the
drop fix the radio?
Answer: The drop allowed positive charge to leak out of the
MOSFET's gate and removing the drop trapped the positive charge on the
gate (where it kept the MOSFET conducting current so that the radio could
operating).
Why: Air is an insulator while water is not. Removing the water
droplet prevented current from flowing away from the gate of the MOSFET
and turning the MOSFET (and the radio) off.
Problem 4: Chapter 14, Case 1a-e (Pg. 519)
Microwave ovens allow for some interesting cooking and funny disasters.
a. Baked Alaska is a dessert in which a hot, baked meringue contains
cold, frozen ice cream. The reverse is Frozen Florida, a dessert in which
cold, frozen meringue contains boiling hot liqueur. Frozen Florida is prepared
by taking a frozen meringue ball (cooked egg whites) containing liquid
liqueur (a water-alcohol mixture that remains liquid at low temperature)
out of the freezer and putting it in a microwave oven briefly. Why does
the liqueur get hot while the meringue remains frozen?
Answer: The liqueur contains liquid water which absorbs microwaves
and becomes hot. The meringue contains frozen, immobile water molecules
which cannot absorb microwaves.
Why: The water molecules in ice cannot turn back and forth to
absorb microwaves and convert their energy into internal energy. Ice and
frozen meringue do not become hot in a microwave oven.
b. If you try to cook an egg in a microwave oven, the egg may explode.
Compare this result to the process of cooking popcorn in a microwave oven.
Answer: The yolk will boil and the increasing pressure inside
the membrane can cause it to explode. In popcorn, water inside the kern
boils and the increasing pressure inside the hull can cause it to explode.
Why: The microwaves heat the water molecules inside the yolk
without overheating and damaging the tough membrane
around the yolk. While conventional cooking usually weakens the membrane
long before it makes the yolk boil, that is
not so with microwave cooking. If the yolk overheats and releases large
numbers of gaseous water molecules, those
water molecules can create an enormous pressure inside the membrane.
When the membrane finally bursts, the steam
throws gooey yolk all over the inside of the oven. Not a pretty sight.
c. Some prepared foods come with browning sheets that contain very
thin metallic layers. These sheets become hot in a microwave oven and help
to brown the surface of the food. Why does a thin metallic film become
so hot?
Answer: The microwaves push currents back and forth in the thin
metallic film, which is a poor conductor and converts much of the electric
energy into thermal energy.
Why: Poor conductors heat up when microwaves drive strong current
back and forth through them. These conductors try to reflect the microwaves
but do a poor job of it. While some of the microwaves reflect, many are
lost as their energy is converted into thermal energy in the poor conductors.
d. If you wrap a piece of food in sturdy metal screening with holes
about 2 millimeters on a side, the food won’t cook in a
microwave oven. Why can’t the microwaves get through the screen’s holes
the way light can?
Answer: Microwaves cannot pass through holes that are significantly
smaller than their wavelengths.
Why: All electromagnetic waves are prevented from passing through
holes in conducting materials if those holes are
much smaller than their wavelengths. Thus an FM radio wave (wavelength
about 3 m), cannot pass well through a
small window in a metal-walled room. On the other hand, light (wavelength
about 0.0000005 m) can pass through a
small window. Microwaves (wavelength about 12 cm), can pass through
a small window but not the metal screening on
it. The holes in the screening are just too small and currents flowing
in the metal screen reflect the microwaves.
e. A microwave oven has air vents and a fan to cool the magnetron
tube. One reason that the tube gets hot is that heat flows out of the electrically
heated cathode at the center of the tube. But there are at least two other
important ways in which the tube’s operation generates heat. What are they?
Answer: (1) Currents flowing back and forth in the copper resonant
cavities lose some of their energies in heating the copper, which is not
a perfect conductor. (2) When the electrons flowing outward from the hot
cathode in the center of the magnetron hit the copper resonant cavities,
they waste some of their energies as thermal energy.
Why: Each of the copper resonant cavities behaves like a tank
circuit, with currents flowing back and forth through it. But copper isn't
a perfect conductor and it wastes some of the current's energy as thermal
energy. The electrons streaming out of the central cathode convert most
of their energies into microwaves by driving current back and forth in
the resonant cavities. But some of their energies is lost when they collide
with the copper cavities and stop. This same sort of impact heating is
what keeps the filaments of fluorescent lamps hot enough to sustain the
discharge through those lamps.
Problem 5: Chapter 14, Case 4a-f (Pg. 520)
When you talk on a cellular telephone, two radio waves connect you with
the telephone network. Your telephone transmits one of these waves and
the other is sent to you by a stationary base unit. Base units are located
a few kilometers apart and each one provides service to the telephones
in its vicinity—its “cell.” A connection between a telephone and a base
unit occupies one of 832 channels currently allocated to cellular communications.
Each channel corresponds to radio waves of two frequencies: a lower frequency
wave (in the range 825.03–849.96 MHz) that’s emitted by the telephone and
a higher frequency wave (in the range 870.03–894.96 MHz) that’s emitted
by the base unit. The waves of different channels are spaced 0.030 MHz
(30 kHz) apart and those of your telephone and its base unit are 45 MHz
apart.
a. When your telephone and its base unit are connected on Channel
1, your telephone’s wave has a frequency of 825.03 MHz and its base unit’s
wave has a frequency of 870.03 MHz. What are the wavelengths of those two
waves?
Answer: 0.3634 meters and 0.3446 meters, respectively.
Why: To find the wavelength of a wave, you divide the speed
of that wave (in this case the speed of light, which is 299,792,458 meters
per second) by the frequency of that wave (in this case, 825,030,000 Hz
and 870,030,000 Hz, respectively). Note that MHz is "megahertz" or "million
Hz" and Hz or "hertz" is equivalent to "per second". In effect, the
speed of the wave tells you how quickly the wave crests spread out and
the frequency of the wave tells you how many wave crests there are. When
you divide wave speed by wave frequency, you are figuring out how tightly
those crests are packed together--their spacing.
b. What is the proper length for the telephone’s antenna if it’s
to have a natural resonance at the frequency of the radio wave it’s transmitting?
Answer: 0.0908 meters and 0.0861 meters, respectively.
Why: A quarter-wavelength antenna is naturally resonant. An
antenna of this length allows the positive and negative charges to flow
to its tip and back exactly one time during each full cycle of the wave.
It takes a quarter cycle for positive charge to reach the tip, a quarter
cycle for that charge to return from the tip, a quarter cycle for negative
charge to reach the tip, and a quarter cycle for that charge to return
from the tip. This then repeats. If the antenna were longer, there wouldn't
be time for the charge to reach the tip and if the antenna were shorter,
the charge would arrive too early.
c. Why will there be a problem if 900 people in a conference center
all try to use their cellular telephones at once?
Answer: The system would run out of available channels.
Why: Each person needs one pair of channels to communicate with
the base unit. Since everyone in the conference center will interact with
the same base unit, those 900 people will need 900 separate channel pairs.
There are only 832 pairs available, so at least 68 people are going to
be out of luck.
d. A base unit’s antenna is oriented vertically because orienting
it horizontally would leave people at certain angles from the base unit
unable to receive its radio wave. Why?
Answer: The radio wave emerging from an antenna is weak or absent
along its axis. With the base unit's antenna horizontal, people located
along the antenna axis won't be able to communicate with the base unit.
Why: When charge accelerates up and down an antenna, it emits
a radio wave. However, this wave is strongest along directions perpendicular
to the charge's motion. If you look at the charge moving on an antenna
from the end of that antenna, you see essentially no motion and can't detect
a radio wave. That's because there basically is no radio wave emitted from
the ends of the antenna. By orienting the antenna vertically, you ensure
that everyone is located perpendicular to the direction of charge motion
on that antenna and can receive a strong signal.
e. Because the base unit’s antenna is vertical, your telephone’s
antenna should also be vertical. Why?
Answer: If the base unit's antenna is vertical, it emits and
absorbs vertically polarized waves. For your antenna to emit and absorb
those waves, too, it must also be vertical.
Why: When charges move up and down a vertical antenna, they
emit radio waves with vertical electric fields. These waves are said to
be vertically polarized. The charges on a vertical antenna only respond
to vertically polarized waves, too, because horizontally polarized waves
would push the charges across the antenna, in the direction they cannot
go.
f. If you’re in a valley or a metal building, your telephone may
lose contact with the base unit. Why?
Answer: The waves go in straight lines and can be blocked when
conducting objects, such as the earth or a metal building, are in the way.
Why: Radio waves go essentially in straight lines, so that if
you cannot see the antenna (more or less) you cannot detect its radio wave.
Why diffraction and scattering effects can help you interact with an antenna
you can't see, the best reception comes when there is an open line-of-sight
between the transmitting and receiving antennas.