Download PHY–309 L. Solutions for Problem set # 3. Non

PHY–309 L.
Solutions for Problem set # 3.
Non-textbook problem I:
(a) In a series circuit, the same current I flows through both resistors while the voltages on
them add up,
Vseries = V1 + V2
= I × R1 + I × R2
(1)
= I × (R1 + R2 ).
Consequently, the net resistance of the series circuit is
Rseries =
Vseries
= R1 + R2 .
I
(2)
For the two resistors in question,
Rseries = 10.0 Ω + 15.0 Ω = 25.0 Ω.
(3)
(b) In a parallel circuit, the same voltage V is applies to both resistors while the currents
through them add up,
Iparallel = I1 + I2
V
V
=
+
R1
R
2
1
1
+
= V ×
.
R2
R2
(4)
Consequently, the net resistance of the parallel circuit is
Rparallel =
V
Iparallel
=
1
1
+
R2
R2
−1
=
R1 R2
.
R1 + R2
(5)
For the two resistors in question,
Rparallel =
10.0 Ω × 15.0 Ω
150 Ω2
=
= 6.00 Ω.
10.0 Ω + 15.0 Ω
25.0 Ω
1
(6)
Non-textbook problem II:
The resistance of a wire is
R =
ρL
A
(7)
where L is the wire’s length, A is its cross-sectional area, and ρ is the resistivity of the
material from which the wire is made. For a round wire,
A = πr2 = π(d/2)2
(8)
where r is the wire’s radius and d = 2r is its diameter. The wire in question has diameter
(not radius!) d = 1.0 mm, so its cross-sectional area is
A = π(1.0 mm/2)2 ≈ 0.78 mm2 = 0.78 · 10−6 m2 .
(9)
Its length is L = 1.6 km = 1.6 · 103 m, while the resistivity of copper is ρ = 17 · 10−9 Ωm.
Therefore, this wire has resistance
R =
(17 · 10−9 Ωm)(1.6 · 103 m)
17 × 1.6
=
Ω ≈ 35 Ω.
−6
2
0.78 · 10 m
0.78
(10)
Textbook problem E.8 at the end of chapter 13:
The internal resistance of a battery is always in series with the load (a light bulb, or a motor,
or whatever) connected to it. The battery in question has internal resistance RI = 1.0 Ω
while the load’s resistance is RL = 9.0 Ω, so the net resistance of the circuit is
Rnet = Rseries = RI + RL = 1.0 Ω + 9.0 Ω = 10.0 Ω.
(11)
(a) The current flowing through the battery and the load is the ratio of the battery’s EMF
E = 3.0 V to the net resistance of the circuit (including the battery’s internal resistance),
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thus
I =
E
3.0 V
=
= 0.30 A.
Rnet
10.0 Ω
(b) The voltage across the load is
V = I × RL = 0.30 A × 9.0 Ω = 2.7 V.
(12)
Note that this voltage is less than the battery’s EMF by
E − V = I × RI = 0.30 A × 1.0 Ω = 0.30 V
(13)
eaten by the battery’s internal resistance.
Textbook problem SP.4 at the end of chapter 13:
Note: to better explain how to analyze such a circuit, I shall treat all the data as having
two significant figures rather than just one, and subsequently calculate everything to two
significant figures.
(a–b) The whole circuit between points A and B is the series connection of 3 blocks: (i) two
parallel resistors on the left; (ii) a single resistor in the middle; (iii) three parallel resistors
on the right. Consequently, the net resistance of the whole circuit is the sum of the 3 block’s
resistances,
Rnet = Rleft + Rmiddle + Rright .
(14)
To find the resistance of the left block, we treat it as a separate parallel circuit, in which
inverse resistances add up:
1
1
1
1
=
+
= 2×
,
Rleft
R1
R2
3.0 Ω
(15)
hence
Rleft =
3.0 Ω
= 1.5 Ω.
2
3
(16)
Similarly, treating the right block as a separate parallel circuit of 3 resistors, we have
1
Rright
=
1
1
1
1
+
+
= 3×
,
R1
R2
R3
3.0 Ω
(17)
hence
Rleft =
3.0 Ω
= 1.0 Ω.
3
(18)
Eqs. (16) and (18) answer part (a) of the problem. To answer part (b) we use eq. (14) for the
series connection of the three blocks. In this equation, we have just found out the resistances
of the left and the right blocks, while the middle block here is just a single resistance, so
Rmiddle = 3 Ω. Consequently,
Rnet = 1.5 Ω + 3.0 Ω + 1.0 Ω = 5.5 Ω.
(19)
(c) Given the voltage between points A and B, the net current through the circuit follows
from its net resistance,
Inet =
Vnet
6.0 V
=
≈ 1.1 A.
Rnet
5.5 Ω
(20)
(d) The whole circuit comprises three blocks connected in series, so the net current through
each block is the whole current through the circuit, I = 1.1 A. In particular, the current
flowing through the middle resistor is the whole 1.1 amperes.
In the left block, the current splits between two parallel resistors. Since they experience
the same voltage and have equal resistances, they carry equal currents. Therefore, each of
the two left resistors carry one half of the net current, (1.1 A)/2 = 0.55 A.
Similarly, in the right block, the current splits between three resistors. Again, since
they have equal resistances they carry equal currents, namely one third of the net current,
(1.1 A)/3 ≈ 0.37 A.
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Textbook problem Q.22 at the end of chapter 13:
As I explained in class, the voltmeter is connected in parallel with the element whose voltage
we want to measure — this way, the voltmeter experiences exactly the same voltage as the
element. But the current through the voltmeter add up to the element’s current, and this
may affect the whole circuit. To minimize the disturbance, the current through the voltmeter
should be much smaller than the element it’s connected to, and since the voltages are the
same, the voltmeter should have much larger resistance than the element.
On the other hand, the ammeter is connected in series with the element so it would
carry exactly the same current as the element. But it’s voltage adds up the the element’s
voltage, which may disturb the whole circuit. To minimize the disturbance, the voltage on
the ammeter should be much smaller than the voltage on the element, and since the currents
are the same, the ammeter should have much smaller resistance than the element.
Altogether, we should have
R[ammeter] R[element] R[voltmeter].
(21)
In particular, the ammeter should have much smaller resistance than the voltmeter.
Textbook problem E.15 at the end of chapter 13:
The electric power consumed by the light bulb is related to the voltage across it and the
current it carries as
P = I × V.
(22)
Given the power of the bulb and the voltage, we can use this formula to find the current
through the bulb as
I =
P
60 W
=
= 0.55 A.
V
110 V
(23)
To be precise, both the 110 V voltage on the bulb and the 0.55 A current through it are
alternating rather than direct. In other words, they vary with time and change directions
120 times a second. The 110 volts is not the maximal voltage across the bulb but rather the
effective average (root(mean square)) voltage. Likewise, the 0.55 A is the effective average
(root(mean square)) current through the bulb.
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Fortunately, the Ohm’s Law works for the effective average (RMS) voltages and currents
in the usual way,
Vrms = R × Irms .
(24)
So to find the resistance of the light bulb, all we need to do is to divide the effective voltage
by the effective current,
R =
110 V
Vrms
=
= 200 Ω.
Irms
0.55 A
(25)
Note that this is the resistance of the hot bulb; a cold bulb would have a lower resistance.
Textbook problem SP.5 at the end of chapter 13:
Note: all currents and voltages in this problem are alternating, so their values (such as
115 V) are effective averages.
(a) The power consumed by each appliance is related to its current and voltage by eq. (22).
Given the common voltage V = 115 V, we can find the current through each appliance from
its power as
Ii =
Pi
.
V
(26)
In particular,
600 W
≈ 5.2 A,
115 V
1200 W
=
≈ 10.4 A,
115 V
500 W
=
≈ 4.4 A.
115 V
Itoaster =
Iiron
Iprocessor
(27)
(b) All the appliances in the house are are connected in parallel. If the toaster, the iron, and
the food processor are all turned on at the same time, their currents would add up to the
net current
Inet = Itoaster + Iiron + Iprocessor ≈ 5.2 A + 10.4 A + 4.4 A ≈ 20 A.
(28)
All this current would flow through the fuse, which is wired in series with the rest of the
circuit. And since the fuse in question is rated for only 15 amperes, it would melt and break
the current.
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PS: Most modern houses in the US are protected by the main circuit breaker rated at 100 to
200 amperes. There are also separate circuit breakers rated at 30 to 60 amperes protecting
each room or a group of rooms, and often there are additional 15 A or 20 A circuit breakers
of fuses protecting each outlet. But older houses or apartments may have more limiting
circuit breakers, such as 60 A for the whole apartment and 15 A for each room. If you live
in such an old apartment, and you need to use several power-hungry appliances at the same
time, try plugging them in in different rooms. Because if you connect them all to the same
low-amperage circuit breaker, it will trip.
(c) When the iron is plugged in to 115 V outlet, its heating element carries current 10.4 A,
see part one. By the Ohm’s law, its resistance is
Riron =
V
Iiron
115 V
≈ 11 Ω.
10.4 A
=
(29)
To be precise, this the heating element’s resistance when it’s hot. When it’s cold, it probably
has a lower resistance.
Textbook question Q.4 at the end of chapter 14:
A bar magnet has a North pole at one and and a South pole at the other end. A more
complicated magnet may have several North poles and several South poles, but there is
always at least one North pole and at least one South pole.
Unlike the electric field lines — which begin at a positive charge and end at a negative
charge — the magnetic field lines do not begin or end but form closed loops. Figure 14.6
(left) on textbook page 285 shows how this work for a bar magnet: Inside the magnet, the
magnetic field lines go from the South pole to the North pole, then they exit the magnet at
the North pole, spread out, loop back, then bunch up and reenter the magnet at the South
pole. But when you consider both the inside and the outside of the magnet, all magnetic
lines form closed loops. The lines do not begin or end at the poles, or anywhere else.
The North pole of a magnet is simply the place where the magnetic field lines exit from
the magnet to the outside space (air, or vacuum, or something else), while the South pole
is the place where the lines re-enter from the outside space back into the magnet. Since the
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lines are closed loops, each line must both exit and re-enter the magnet, so there mush be
both a North pole and a South pole.
So what happens when you break a bar magnet in half? Naively, one might expect the
each half to have just one pole, say the North pole in the top half and the South pole in
the bottom half, but that’s not how the magnets work. Instead, in addition to the North
pole the top half ‘inherits’ from the original magnet, it also gets a South pole at the place
where it broke off. Likewise, in addition to the South pole the bottom half ‘inherits’ from
the original magnet, it also gets a North pole at the breaking point. Thus, each half becomes
a complete magnet with both a North pole and a South pole.
The bottom line is, any magnet has both a North pole and a South pole. Magnets with just
one pole — either North or South — do not exist.
PS: To be precise, all known kinds of magnets have both a North pole and a South pole. You
cannot make a magnet with just one pole from any kind of electric currents or any know kind
of matter. Even the magnetic fields of elementary particles — or rather all the elementary
particles discovered up to now — have field lines closed in loops, just like magnets with both
a North pole and a South pole.
But according to modern theoretical physics, there might exist so-far-undiscovered elementary particles with just one magnetic pole, either North or South; such hypothetical
particles are called magnetic monopoles. Many physicists have tried to find such monopoles
among the cosmic rays, or to make them in an accelerator lab, but so far there were no
discoveries. So if magnetic monopoles exist at all, they must be extremely rare, no more
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than one monopole per 50 billion cubic miles of space, perhaps as rare as one monopole per
Galaxy, or even rarer.
Thus, any magnet you would ever see in your life is going to have both a North pole and
a South pole.
Textbook question Q.5 at the end of chapter 14:
In a word, NO.
In the absence of nearby electric currents or magnets — or pieces of iron or other magnetic
materials — the compass needle lines up with the geomagnetic field, i.e., the magnetic field
generated by the planet Earth. This field is directed roughly from the South to the North,
but not exactly. So the northern end of the compass needle points roughly to the North, but
not exactly to the geographic North pole on the Earth’s rotation axis.
Since the magnetic field of the Earth is more complicated than the field of a bar magnet,
locating the Earth’s magnetic poles is somewhat tricky. Also, they do not stay fixed in their
?
places but keep moving a few miles every year, please visit this web site. The exact direction
of the compass needle depends on where you are: In general, the northern end of the needle
points to somewhere in the Canadian Arctic while the southern end points to somewhere in
the Eastern Antarctica. The angle between the compass direction and the geographic North
†
is called the magnetic declination, please visit this wikipedia article for more information.
The declination varies from place to place, and navigators have been making declination
maps since the 18th century.
? http://www.geomag.bgs.ac.uk/education/poles.html
† http://en.wikipedia.org/wiki/Magnetic declination
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