week 6 - university of california, san...

week 6chapter 31

Current and Resistance

Which is the correct way to light

the lightbulb with the battery?4) all are correct

5) none are correct

1) 3)2)

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).

Which is the correct way to light

the lightbulb with the battery?4) all are correct

5) none are correct

1) 3)2)

Why does the light in a room come on instantly when you flip a switch several meters away?

1. Electrons travel at the speed of light through the wire.

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.

Why does the light in a room come on instantly when you flip a switch several meters away?

Recall water analogy

1. Electrons travel at the speed of light through the wire.

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.

These four wires are made of the same metal. Rank in order, from largest to smallest, the electron currents ia to id.

1. id > ia > ib > ic

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.

(Note: electron current is number ofelectrons moving across the wire section per unit time)

1. ic > ie > ia > ib = id

2. id > ib > ie > ia = ic

3. ic = id > ie > ia = ib

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

Rank in order, from largest to smallest, the current densities J

a to J

d in these four wires.

1. Jc > Jb > Ja > Jd

2. Jb > Ja = Jd > Jc

3. Jb > Ja > Jc > Jd

4. Jc > Jb > Ja = Jd

5. Jb = Jd > Ja > Jc

Rank in order, from largest to smallest, the current densities J

a to J

d in these four wires.

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

1. Jc > Jb > Ja > Jd

2. Jb > Ja = Jd > Jc

3. Jb > Ja > Jc > Jd

4. Jc > Jb > Ja = Jd

5. Jb = Jd > Ja > Jc

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


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,

Recall J = ne e vd, and J = I/AMoving from larger to smaller diameter implies moving from smaller to larger Jso 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



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


Recall, again, J = ne e vd, and J = I/AThroughout the resistor and in the wires on either endthe current is I and since the diameter of the wireis the same before entering and after leaving the resistorthe current density J is the same in both wires, and hence theelectrons have the same drift velocity


Compared to the potential energy of an electron before entering the resistor, the potential energy of an electron after leaving the resistor is

1. greater

2. less

3. the same



Compared to the potential energy of an electron before entering the resistor, the potential energy of an electron after leaving the resistor is

1. greater

2. less

3. the same


On general grounds: energy is dissipated in the resistorso the charge carriers (electrons) must loose energy.

In equations, the electric potential decreases from one end of the resistor to the other in the directionof the current. Electrons move opposite the current, so theymove 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 is0.085 V/m. What is the wire resistivity?

Ans: 5.0 x 10-8 Ω⋅m

The two 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 two segments of the wire?

Ans: 1/2

What are the magnitude and the direction of the current in the fifth wire?

1. 15 A into the junction

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

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.

You double the voltage across a

certain conductor and you

observe the current increases

three times. What can you

conclude?

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

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.

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

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) it decreases by a factor 4


3) it stays the same

4) it increases by a factor 2


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.

A wire of resistance R is stretched

uniformly (keeping its volume

constant) until it is twice its original

length. What happens to the

resistance?



3) it stays the same



Conductors a to d are all made of the same material. Rank in order, from largest to smallest, the resistances Ra to Rd.

1. Ra > Rc > Rb > Rd

2. Rb > Rd > Ra > Rc

3. Rc > Ra > Rd > Rb

4. Rc > Ra = Rd > Rb

5. Rd > Rb > Rc > Ra

Conductors a to d are all made of the same material. Rank in order, from largest to smallest, the resistances Ra to Rd.

1. Ra > Rc > Rb > 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 Ib. current density Jc. electric field Ed. drift velocity vd

e. mean time between collisions τf. electron current i

Ans: Itop = Imid = Ibot = 10 AJtop = Jbot = 3.18x106 A/m2, Jmid = 1.27x107 A/m2

Etop = Ebot = 0.0909 V/m, Emid = 0.364 V/mτ = 2.07x10-14 sitop = ibot = imid = 6.25x1019 s-1

σ = 3.5 x 107 Ω-1 m-1

n = 6.0 x 1028 m-3

When you rotate the knob of a light

dimmer, what is being changed in

the electric circuit?

1) the power

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.

When you rotate the knob of a light

dimmer, what is being changed in

the electric circuit?

1) the power

2) the current

3) the voltage

4) both (1) and (2)

5) both (2) and (3)

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



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


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 emfthe current is I and since the diameter of the wireis the same before entering and after leaving the source of emfthe current density J is the same in both wires, and hence theelectrons have the same drift velocity


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



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


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 two 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 between 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.

! = "0

!1#2! 1

#1

"I

AAns:

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)

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