Physics 106 - How Things Work II - Spring 2008
Problem Set #3 — with solutions
1. There are 50 turns in the primary coil of a transformer. The same transformer has
250 turns in the secondary coil.
a. If 12 Volts AC is applied to the primary coil, what will be the voltage coming out
of the secondary coil?
Here we recall the relation we saw during lecture 16:
# of turns in primary coil
voltagein
=
voltageout
# of turns in secondary coil
Inserting the values above we have:
50 turns
1
12 V
=
=
voltageout
250 turns
5
Rearranging, we have:
voltageout = 5 · 12 V = 60 V
b. A lightbulb is attached to the secondary coil while (as was the case in the previous
part) 12 Volts AC is applied to the primary coil. If the current in the primary
coil is 5 A, how much current flows through the secondary coil?
Here we recall another relationship shown during lecture 16 regarding transformers:
Powerin = Powerout
We also recall that Powerin = currentin · voltagein , and similarly for Powerout .
Recalling that from part “a” we have determined the output voltage to be 60 V,
we have that:
currentin · voltagein = currentout · voltageout
(5 A) · (12 V) = (currentout ) · (60 V)
Rearranging, we have:
currentout =
(5 A) · (12 V)
= 1A
60 V
c. If 12 Volts DC is applied to the primary coil, what will be the voltage coming
out of the secondary coil?
A transformer requires AC current to operate. Otherwise the primary coil does
not produce a changing magnetic field, and only a changing magnetic field can
induce current in the secondary coil. If 12 V DC is fed into the transformer, the
output voltage will be zero.
Physics 106 - How Things Work II - Spring 2008
Problem Set #3 with Solutions — continued
2. The rear defroster of your car operates on a current of 5A. If the voltage drop across
it is 10V, how much power is it consuming as it melts the frost?
The power consumed by a load can be computed using the relation P = I V. So in
this case we have:
P = 5 A · 10 V = 50 Watts .
3. A 2500 Ω heating filament is subjected to a voltage drop of 100 V.
a. How much current will flow through it?
Here we use Ohm’s law: E = I R. For this problem we need to solve for the
current, so we have:
100 V
E
=
= 0.04 A
I=
R
2500 Ω
b. Using your answer to part (a), how much power will the heating element consume?
Using P = I V we have that:
P = 0.04A · 100 V = 4 Watts .
4. One type of microphone has a permanent magnet and a coil of wire that move relative
to one another in response to sound waves. Why is the current in the coil related to
the motion?
When a magnet moves closer to or further from a coil of wire, the coil of wire
sees a changing magnetic field, which produces a voltage. If the magnet and the
coil move with respect to one another in response to sound waves, the voltage in
the coil will be related to the sound, as we would expect in a microphone.
5. A house is equipped with a lightening rod and a positively charged cloud passes
overhead.
a. What type of charge (positive or negative) accumulates on the lightning rod?
Negative charge is attracted to the positively charged cloud overhead, and would
thus accumulate on the lightning rod.
b. How is it that the lightning rod decreases the chance that the house will be struck
by lightning?
As is discussed in a “side bar” on page 322 of our text, a lightning rod functions
by producing corona discharge (in this case negative) that will diminish any local
buildup of electric charge. In this way the electric fields are diminished around
the house, which could otherwise lead to a lightning strike.
Physics 106 - How Things Work II - Spring 2008
Problem Set #3 with Solutions — continued
6. On electric motors.
a. You cannot make an electric motor using direct current and electromagnets without using switches. Why?
If we imagine a permanent bar magnet spinning around between the pole faces of
a fixed permanent magnet (much as is depicted in one of the slides from lecture
18), all that happens is that the bar magnet aligns itself so that the north pole of
the spinning magnet is close to the south pole of the fixed magnet, and similarly
for the south pole of the spinning magnet. This is basically what happens when
a compass needle comes to rest and points north. To have a motor, something
needs to cause the poles of either the spinning or the fixed magnet to periodically
“switch polarity” (that is, for a north pole to become a south pole and a south pole
to become a north pole). If you use DC current and no switches, you essentially
have two permanent magnets, so by the argument above, the motor will not spin.
b. Explain why an AC synchronous motor always spins at either 60 Hz or at an
integer fraction of that rate if the rotor has multiple pairs of poles.
In an AC synchronous motor, the thing that is causing one of the magnets to
switch polarity is the AC current itself. If the rotor has two poles, this ensures
that the rotor will spin at exactly the rate at which the line current is cycling,
which in the U.S. is 60 Hz. It is possible, however, to have a rotor with 4 poles,
or 6 poles, etc. In this case, the time it takes for the “correct pole” to line up
with the pole face of the fixed magnet will be 1/2, 1/3, etc. of the time needed if
the rotor were essentially the equivalent of a simple bar magnet. Thus, the rate at
which the rotor would spin would be 1/2, 1/3, etc. of what it would be otherwise,
i.e. 30 Hz, 20 Hz, etc.
7. When you swipe a credit or debit card on a card reader, why is it important to do so
relatively quickly?
The strip on the credit card can be thought of as being made up by a series of little
magnets. Sometimes the north poles are facing upwards and sometimes the south
poles are facing upwards. The card reader will contain some type of small coil.
When we swipe the credit card, the coil sees a changing magnet field corresponding
to the pattern on the credit card. The changing magnetic field, in turn, produces
voltages in the coil that correspond to the pattern on the credit card. The faster
the magnetic fields are changing that are seen by the reader, the larger the voltages
that are induced in the coil. If the voltages induced in the reader are not large
enough, they are not big enough to be detected.