Download d - UCSD Physics

week 6
chapter 31
Current and Resistance
Which is the correct way to light
the lightbulb with the battery?
1)
4) all are correct
5) none are correct
2)
3)
Which is the correct way to light
the lightbulb with the battery?
1)
4) all are correct
5) none are correct
2)
3)
Current can only flow if there is a continuous connection from
the negative terminal through the bulb to the positive terminal.
This is only the case for Fig. (3).
1. Electrons travel at the speed of light through the wire.
Why does the
light in a room
come on
instantly when
you flip a
switch several
meters away?
2. Because the wire between the switch and the bulb is
already full of electrons, a flow of electrons from the
switch into the wire immediately causes electrons to flow
from the other end of the wire into the lightbulb.
3. The switch sends a radio signal which is received by a
receiver in the light which tells it to turn on.
4. Optical fibers connect the switch with the light, so the
signal travels from switch to the light at the speed of
light in an optical fiber.
1. Electrons travel at the speed of light through the wire.
Why does the
light in a room
come on
instantly when
you flip a
switch several
meters away?
2. Because the wire between the switch and the bulb is
already full of electrons, a flow of electrons from the
switch into the wire immediately causes electrons to flow
from the other end of the wire into the lightbulb.
3. The switch sends a radio signal which is received by a
receiver in the light which tells it to turn on.
4. Optical fibers connect the switch with the light, so the
signal travels from switch to the light at the speed of
light in an optical fiber.
Recall water analogy
1. id > ia > ib > ic
These four wires are made of the
same metal. Rank in order, from
largest to smallest, the electron
currents ia to id.
2. ib = id > ia = ic
3. ic > ib > ia > id
4. ic > ia = ib > id
5. ib = ic > ia = id
1. id > ia > ib > ic
These four wires are made of the
same metal. Rank in order, from
largest to smallest, the electron
currents ia to id.
2. ib = id > ia = ic
3. ic > ib > ia > id
4. ic > ia = ib > id
5. ib = ic > ia = id
The two charged rings are a model of
the surface charge distribution along a
wire. Rank in order, from largest to
smallest, the electron currents ia to ie at
the midpoint between the rings.
1.
ic > ie > ia > ib = id
2.
id > ib > ie > ia = ic
3.
ic = id > ie > ia = ib
(Note: electron current is number of
electrons moving across the wire
section per unit time)
4.
ib = id > ia = ic = ie
5.
ia = ib > ie > ic = id
The two charged rings are a model of
the surface charge distribution along a
wire. Rank in order, from largest to
smallest, the electron currents ia to ie at
the midpoint between the rings.
1.
ic > ie > ia > ib = id
2.
id > ib > ie > ia = ic
3.
ic = id > ie > ia = ib
(Note: electron current is number of
electrons moving across the wire
section per unit time)
4.
ib = id > ia = ic = ie
5.
ia = ib > ie > ic = id
Electrons flow through a 1.6-mm-diameter aluminum wire at 2.0 x 10-4 m/s. How
many electrons move through a cross section of the wire each day?
Ans: 2.1 x 1024 electrons
(used n= 6.0 x 1028 m-3 for aluminum, from table 31.1)
The current in an electric dryer is 10.0 A. How much charge and how many
electrons flow through the hair dryer in 5.0 min
Ans: 3.0 x 103C, 1.88 x 1022
1. Jc > Jb > Ja > Jd
Rank in order, from largest to
smallest, the current densities
Ja to Jd in these four wires.
2. Jb > Ja = Jd > Jc
3. Jb > Ja > Jc > Jd
4. Jc > Jb > Ja = Jd
5. Jb = Jd > Ja > Jc
1. Jc > Jb > Ja > Jd
2. Jb > Ja = Jd > Jc
Rank in order, from largest to
smallest, the current densities
Ja to Jd in these four wires.
3. Jb > Ja > Jc > Jd
4. Jc > Jb > Ja = Jd
5. Jb = Jd > Ja > Jc
J = I/A, does not depend on conductivity σ.
Ja = I/πr2
(2I)/πr2 = 2Ja
(2I)/π(2r)2 = 1/2 Ja
I/πr2 = Ja
Two copper wires of different
diameter are joined end-to-end,
and a current flows in the wire
combination.
When electrons move from the
larger-diameter wire into the
smaller-diameter wire,
1. their drift speed increases
2. their drift speed decreases
3. their drift speed stays the same
4. not enough information given to decide
Two copper wires of different
diameter are joined end-to-end,
and a current flows in the wire
combination.
When electrons move from the
larger-diameter wire into the
smaller-diameter wire,
1. their drift speed increases
2. their drift speed decreases
3. their drift speed stays the same
4. not enough information given to decide
Recall J = ne e vd, and J = I/A
Moving from larger to smaller diameter
implies moving from smaller to larger J
so J increases and hence drift velocity increases too
Electrons in an electric circuit pass
through a resistor. The wire has the
same diameter on each side of the
resistor.
Compared to the drift speed of the
electrons before entering the resistor,
the drift speed of the electrons after
leaving the resistor is
1. faster
2. slower
3. the same
4. not enough information
given to decide
Electrons in an electric circuit pass
through a resistor. The wire has the
same diameter on each side of the
resistor.
Compared to the drift speed of the
electrons before entering the resistor,
the drift speed of the electrons after
leaving the resistor is
1. faster
2. slower
3. the same
4. not enough information
given to decide
Recall, again, J = ne e vd, and J = I/A
Throughout the resistor and in the wires on either end
the current is I and since the diameter of the wire
is the same before entering and after leaving the resistor
the current density J is the same in both wires, and hence the
electrons have the same drift velocity
Electrons in an electric circuit pass
through a resistor. The wire has the same
diameter on each side of the resistor.
1. greater
Compared to the potential energy of an
electron before entering the resistor, the
potential energy of an electron after
leaving the resistor is
3. the same
2. less
4. not enough information
given to decide
Electrons in an electric circuit pass
through a resistor. The wire has the same
diameter on each side of the resistor.
1. greater
Compared to the potential energy of an
electron before entering the resistor, the
potential energy of an electron after
leaving the resistor is
3. the same
2. less
4. not enough information
given to decide
On general grounds: energy is dissipated in the resistor
so the charge carriers (electrons) must loose energy.
In equations, the electric potential decreases from one end of the
resistor to the other in the direction
of the current. Electrons move opposite the current, so they
move towards higher electric potential V. The potential energy
of electrons is U = −eV, it decreases when V increases.
A 3.0-mm-diameter wire carries a 12 A current when the electric field is
0.085 V/m. What is the wire resistivity?
Ans: 5.0 x 10-8 Ω⋅m
The t wo segments of the wire in the figure have equal diameters but different
conductivities σ1 and σ2. Current I passes through this wire. If the
conductivities have the ratio σ2/σ1 =2, what is the ratio E2/E1 of the
electric field strengths in the t wo segments of the wire?
Ans: 1/2
1. 15 A into the junction
What are the magnitude and the
direction of the current in the
fifth wire?
2. 15 A out of the junction
3. 1 A into the junction
4. 1 A out of the junction
5. Not enough data to determine
1. 15 A into the junction
What are the magnitude and the
direction of the current in the
fifth wire?
2. 15 A out of the junction
3. 1 A into the junction
4. 1 A out of the junction
5. Not enough data to determine
A wire connects the positive
and negative terminals of a
battery. Two identical wires
connect the positive and
negative terminals of an
identical battery. Rank in order,
from largest to smallest, the
currents Ia to Id at points a to d.
1. Ic = Id > Ia > Ib
2. Ia = Ib > Ic = Id
3. Ic = Id > Ia = Ib
4. Ia = Ib = Ic = Id
5. Ia > Ib > Ic = Id
A wire connects the positive
and negative terminals of a
battery. Two identical wires
connect the positive and
negative terminals of an
identical battery. Rank in order,
from largest to smallest, the
currents Ia to Id at points a to d.
1. Ic = Id > Ia > Ib
2. Ia = Ib > Ic = Id
3. Ic = Id > Ia = Ib
4. Ia = Ib = Ic = Id
5. Ia > Ib > Ic = Id
You double the voltage across a
certain conductor and you
observe the current increases
three times. What can you
conclude?
1) Ohm’s law is obeyed since the
current still increases when V
increases
2) Ohm’s law is not obeyed
3) This has nothing to do with Ohm’s
law
You double the voltage across a
certain conductor and you
observe the current increases
three times. What can you
conclude?
1) Ohm’s law is obeyed since the
current still increases when V
increases
2) Ohm’s law is not obeyed
3) This has nothing to do with Ohm’s
law
Ohm’s law, V = I R, states that the
relationship between voltage and
current is linear. Thus for a conductor
that obeys Ohm’s Law, the current must
double when you double the voltage.
Follow-up: Where could this situation occur?
Two wires, A and B, are made of the
same metal and have equal length, but
the resistance of wire A is four times
the resistance of wire B. How do their
diameters compare?
1) dA = 4 dB
2) dA = 2 dB
3) dA = dB
4) dA = 1/2 dB
5) dA = 1/4 dB
Two wires, A and B, are made of the
same metal and have equal length, but
the resistance of wire A is four times
the resistance of wire B. How do their
diameters compare?
1) dA = 4 dB
2) dA = 2 dB
3) dA = dB
4) dA = 1/2 dB
5) dA = 1/4 dB
The resistance of wire A is greater because its area is less than
wire B. Since area is related to radius (or diameter) squared,
the diameter of A must be two times less than B.
A wire of resistance R is stretched
uniformly (keeping its volume
constant) until it is twice its original
length. What happens to the
resistance?
1)
2)
3)
4)
5)
it decreases by a factor 4
it decreases by a factor 2
it stays the same
it increases by a factor 2
it increases by a factor 4
A wire of resistance R is stretched
uniformly (keeping its volume
constant) until it is twice its original
length. What happens to the
resistance?
1)
2)
3)
4)
5)
it decreases by a factor 4
it decreases by a factor 2
it stays the same
it increases by a factor 2
it increases by a factor 4
Keeping the volume (= area x length) constant means
that if the length is doubled, the area is halved.
Since
, this increases the resistance by four.
1. Ra > Rc > Rb > Rd
Conductors a to d are all made
of the same material. Rank in
order, from largest to smallest,
the resistances Ra to Rd.
2. Rb > Rd > Ra > Rc
3. Rc > Ra > Rd > Rb
4. Rc > Ra = Rd > Rb
5. Rd > Rb > Rc > Ra
1. Ra > Rc > Rb > Rd
Conductors a to d are all made
of the same material. Rank in
order, from largest to smallest,
the resistances Ra to Rd.
2. Rb > Rd > Ra > Rc
3. Rc > Ra > Rd > Rb
4. Rc > Ra = Rd > Rb
5. Rd > Rb > Rc > Ra
L/r2
L/(2r)2=(1/4) L/r2
(2L)/(r)2= 2 L/r2
(2L)/(2r)2=(1/2) L/r2
An aluminum wire consists of the three segments shown in the figure. The
current in the top segment is 10 A. For each of these segments find the:
a. current I
b. current density J
c. electric field E
d. drift velocity vd
e. mean time bet ween collisions τ
f. electron current i
σ = 3.5 x 107 Ω-1 m-1
n = 6.0 x 1028 m-3
Ans: Itop = Imid = Ibot = 10 A
Jtop = Jbot = 3.18x106 A/m2, Jmid = 1.27x107 A/m2
Etop = Ebot = 0.0909 V/m, Emid = 0.364 V/m
τ = 2.07x10-14 s
itop = ibot = imid = 6.25x1019 s-1
1) the power
When you rotate the knob of a light
dimmer, what is being changed in
the electric circuit?
2) the current
3) the voltage
4) both (1) and (2)
5) both (2) and (3)
1) the power
When you rotate the knob of a light
dimmer, what is being changed in
the electric circuit?
2) the current
3) the voltage
4) both (1) and (2)
5) both (2) and (3)
The voltage is provided at 120 V from the
outside. The light dimmer increases the
resistance and therefore decreases the current
that flows through the lightbulb.
Follow-up: Why does the voltage not change?
Electrons in an electric circuit pass
through a source of emf. The wire has
the same diameter on each side of the
source of emf.
Compared to the drift speed of the
electrons before entering the source
of emf, the drift speed of the electrons
after leaving the source of emf is
1. faster
2. slower
3. the same
4. not enough information
given to decide
Electrons in an electric circuit pass
through a source of emf. The wire has
the same diameter on each side of the
source of emf.
Compared to the drift speed of the
electrons before entering the source
of emf, the drift speed of the electrons
after leaving the source of emf is
1. faster
2. slower
3. the same
4. not enough information
given to decide
Recall, again, J = ne e vd, and J = I/A.
This is just as in the case with a resistor, done earlier.
In the wires on either end of the source of emf
the current is I and since the diameter of the wire
is the same before entering and after leaving the source of emf
the current density J is the same in both wires, and hence the
electrons have the same drift velocity
Electrons in an electric circuit pass
through a source of emf. The wire has
the same diameter on each side of the
source of emf.
Compared to the potential energy of an
electron before entering the source of
emf, the potential energy of an electron
after leaving the source of emf is
1. greater
2. less
3. the same
4. not enough information
given to decide
Electrons in an electric circuit pass
through a source of emf. The wire has
the same diameter on each side of the
source of emf.
Compared to the potential energy of an
electron before entering the source of
emf, the potential energy of an electron
after leaving the source of emf is
1. greater
2. less
3. the same
4. not enough information
given to decide
On general grounds: source of emf does work on
the charge carriers (electrons) which therefore must gain energy.
In equations, the electric potential increases across the source of emf
in the direction of the current. Electrons move opposite the current, so
they move towards lower electric potential V. The potential energy
of electrons is U = −eV, it increases when V decreases.
The figure shows a wire that is made of t wo equal diameter segments with
conductivities σ1 and σ2. When a current I passes through the wire, a thin layer of
charge appears at the boundary bet ween the segments.
Find an expression for the surface charge density η on the boundary. Give your
result in terms of I, σ1, σ2, and the wire’s cross-section area A.
Ans:
η = "0
!
1
1
−
σ2
σ1
"
I
A
Example: What should be the diameter of an aluminum wire that carries 15 A
when the voltage across 1.0 m of the wire is 0.25 V?
Ans: 1.4 mm
(need resistivity of aluminum: 2.65 x 10-8 Ω.m)