chapter 25 electric currents and resistancepeople.virginia.edu/~ben/2415131/lecture_8.pdf ·...

Copyright © 2009 Pearson Education, Inc.

Chapter 25

Electric Currents and

Resistance


25-4 Resistivity

Example 25-5: Speaker wires.

Suppose you want to connect

your stereo to remote

speakers. (a) If each wire must

be 20 m long, what diameter

copper wire should you use to

keep the resistance less than

0.10 Ω per wire? (b) If the

current to each speaker is 4.0

A, what is the potential

difference, or voltage drop,

across each wire?


For any given material, the resistivity

increases with temperature:

Semiconductors are complex materials, and

may have resistivities that decrease with

temperature.

25-4 Resistivity

0 0

0 0

1

1

T T T

R T R T T


25-4 Resistivity

Example 25-7: Resistance thermometer.

The variation in electrical resistance with

temperature can be used to make precise

temperature measurements. Platinum is

commonly used since it is relatively free from

corrosive effects and has a high melting point.

Suppose at 20.0°C the resistance of a platinum

resistance thermometer is 164.2 Ω. When

placed in a particular solution, the resistance is

187.4 Ω. What is the temperature of this

solution? 0.003927Pt m

ConcepTest 25.3a Wires I

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 = 4dB

2) dA = 2dB

3) dA = dB

4) dA = 1/2dB

5) dA = 1/4dB

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 the diameter of B.

ConcepTest 25.3a Wires I

R ρA

,

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 = 4dB

2) dA = 2dB

3) dA = dB

4) dA = 1/2dB

5) dA = 1/4dB


Power, as in kinematics, is the energy

transformed by a device per unit time:

25-5 Electric Power

or


The unit of power is the watt, W.

For ohmic devices, we can make the

substitutions:

25-5 Electric Power


25-5 Electric Power

Example 25-8: Headlights.

Calculate the resistance of a 40-W

automobile headlight designed for 12 V.


What you pay for on your electric bill is

not power, but energy – the power

consumption multiplied by the time.

We have been measuring energy in

joules, but the electric company

measures it in kilowatt-hours, kWh:

1 kWh = (1000 W)(3600 s) = 3.60 x 106 J.

25-5 Electric Power


25-5 Electric Power

Example 25-9: Electric heater.

An electric heater draws a steady 15.0

A on a 120-V line. How much power

does it require and how much does it

cost per month (30 days) if it operates

3.0 h per day and the electric company

charges 9.2 cents per kWh?


Current from a battery

flows steadily in one

direction (direct current,

DC). Current from a

power plant varies

sinusoidally (alternating

current, AC).

25-7 Alternating Current


The voltage varies sinusoidally with time:

as does the current:


, ,


Multiplying the current and the voltage gives

the power:



Usually we are interested in the average power:


.


The current and voltage both have average

values of zero, so we square them, take the

average, then take the square root, yielding the

root-mean-square (rms) value:




Example 25-13: Hair dryer.

(a) Calculate the resistance and the peak current

in a 1000-W hair dryer connected to a 120-V line.

(b) What happens if it is connected to a 240-V line

in Britain?


Electrons in a conductor have large, random

speeds just due to their temperature. When a

potential difference is applied, the electrons

also acquire an average drift velocity, which is

generally considerably smaller than the

thermal velocity.

25-8 Microscopic View of Electric

Current: Current Density and Drift

Velocity




Velocity

We define the current density (current per

unit area) – this is a convenient concept

for relating the microscopic motions of

electrons to the macroscopic current:

If the current is not uniform:

. (Remember the

water in the pipe)


This drift speed is related to the current in the

wire, and also to the number of electrons per unit

volume:



Velocity

and




Velocity

Example 25-14: Electron speeds in a wire.

A copper wire 3.2 mm in diameter carries a 5.0-

A current. Determine (a) the current density in

the wire, and (b) the drift velocity of the free

electrons. (c) Estimate the rms speed of

electrons assuming they behave like an ideal

gas at 20°C. Assume that one electron per Cu

atom is free to move (the others remain bound

to the atom).




Velocity

The electric field inside a current-carrying

wire can be found from the relationship

between the current, voltage, and resistance.

Writing R = ρ l/A, I = jA, and V = El , and

substituting in Ohm’s law gives:




Velocity

Example 25-15: Electric field inside a wire.

What is the electric field inside the wire of

the earlier example? (The current density

was found to be 6.2 x 105 A/m2.)


In general, resistivity

decreases as

temperature decreases.

Some materials,

however, have

resistivity that falls

abruptly to zero at a

very low temperature,

called the critical

temperature, TC.

25-9 Superconductivity

Purely quantum mechanical; CANNOT be

explained using classical physics.


Experiments have shown that currents, once

started, can flow through these materials for

years without decreasing even without a

potential difference.

Critical temperatures are low; for many years no

material was found to be superconducting above

23 K.

Since 1987, new materials have been found that

are superconducting below 90 K, and work on

higher temperature superconductors is

continuing.

25-9 Superconductivity


• A battery is a source of constant potential

difference.

• Electric current is the rate of flow of electric

charge.

• Conventional current is in the direction that

positive charge would flow.

• Resistance is the ratio of voltage to current:

Summary of Chapter 25


• Ohmic materials have constant resistance,

independent of voltage.

• Resistance is determined by shape and

material:

• ρ is the resistivity.



• Power in an electric circuit:

• Direct current is constant.

• Alternating current varies sinusoidally:



• The average (rms) current and voltage:

• Relation between drift speed and current:



Chapter 26

DC Circuits


Electric circuit needs battery or generator to

produce current – these are called sources of

emf.

Battery is a nearly constant voltage source, but

does have a small internal resistance, which

reduces the actual voltage from the ideal emf:

26-1 EMF and Terminal Voltage


This resistance behaves as though it were in

series with the emf.

26-1 EMF and Terminal Voltage


26-1 EMF and Terminal Voltage Example 26-1: Battery with internal resistance.

A 65.0-Ω resistor is

connected to the

terminals of a battery

whose emf is 12.0 V and

whose internal

resistance is 0.5 Ω.

Calculate (a) the current

in the circuit, (b) the

terminal voltage of the

battery, Vab, and (c) the

power dissipated in the

resistor R and in the

battery’s internal resistance r.


A series connection has a single path from

the battery, through each circuit element in

turn, then back to the battery.

26-2 Resistors in Series and in

Parallel


The current through each resistor is the

same; the voltage depends on the

resistance. The sum of the voltage

drops across the resistors equals the

battery voltage:

26-2 Resistors in Series

1 2 3 1 2 3

1 2 3Series

eq

V V V V IR IR IR

I R R R IR


From this we get the equivalent resistance (that

single resistance that gives the same current in

the circuit):

26-2 Resistors in Series

Unless an internal

resistance r is

specified assume V

constant.

ConcepTest 26.1a Series Resistors I

9 V

Assume that the voltage of the battery

is 9 V and that the three resistors are

identical. What is the potential

difference across each resistor?

1) 12 V

2) zero

3) 3 V

4) 4 V

5) you need to know the

actual value of R

Since the resistors are all equal,

the voltage will drop evenly

across the 3 resistors, with 1/3 of

9 V across each one. So we get a

3 V drop across each.

ConcepTest 26.1a Series Resistors I

9 V

Assume that the voltage of the battery

is 9 V and that the three resistors are

identical. What is the potential

difference across each resistor?

1) 12 V

2) zero

3) 3 V

4) 4 V

5) you need to know the

actual value of R

Follow-up: What would be the potential difference if

R = 1 , 2 , 3 ?

ConcepTest 26.1b Series Resistors II

12 V

R1 = 4 R2 = 2

In the circuit below, what is the

voltage across R1?

1) 12 V

2) zero

3) 6 V

4) 8 V

5) 4 V

ConcepTest 26.1b Series Resistors II

12 V

R1 = 4 R2 = 2


voltage across R1?

1) 12 V

2) zero

3) 6 V

4) 8 V

5) 4 V

The voltage drop across R1 has

to be twice as big as the drop

across R2. This means that V1 =

8 V and V2 = 4 V. Or else you

could find the current I = V/R =

(12 V)/(6 = 2 A, and then use

Ohm’s law to get voltages.

Follow-up: What happens if the voltage is doubled?


A parallel connection splits the current; the

voltage across each resistor is the same:

26-2 Resistors in Parallel


The total current is the sum of the currents

across each resistor:


,


This gives the reciprocal of the equivalent

resistance:



An analogy using water

may be helpful in

visualizing parallel

circuits. The water

(current) splits into two

streams; each falls the

same height, and the total

current is the sum of the

two currents. With two

pipes open, the resistance

to water flow is half what

it is with one pipe open.




Parallel Conceptual Example 26-2: Series or parallel?

(a) The lightbulbs in the figure are identical.

Which configuration produces more light? (b)

Which way do you think the headlights of a car

are wired? Ignore change of filament resistance R

with current.

ConcepTest 26.2a Parallel Resistors I


current through R1?

10 V

R1 = 5

R2 = 2

1) 10 A

2) zero

3) 5 A

4) 2 A

5) 7 A

The voltage is the same (10 V) across each

resistor because they are in parallel. Thus,

we can use Ohm’s law, V1 = I1R1 to find the

current I1 = 2 A.

ConcepTest 26.2a Parallel Resistors I


current through R1?

10 V

R1 = 5

R2 = 2

1) 10 A

2) zero

3) 5 A

4) 2 A

5) 7 A

Follow-up: What is the total current through the battery?

ConcepTest 26.2b Parallel Resistors II

1) increases

2) remains the same

3) decreases

4) drops to zero

Points P and Q are connected to a

battery of fixed voltage. As more

resistors R are added to the parallel

circuit, what happens to the total

current in the circuit?

ConcepTest 26.2b Parallel Resistors II

1) increases

2) remains the same

3) decreases

4) drops to zero

As we add parallel resistors, the overall

resistance of the circuit drops. Since V =

IR, and V is held constant by the battery,

when resistance decreases, the current

must increase.

Points P and Q are connected to a

battery of fixed voltage. As more

resistors R are added to the parallel

circuit, what happens to the total

current in the circuit?

Follow-up: What happens to the current through each resistor?



Parallel Conceptual Example 26-3: An illuminating surprise.

A 100-W, 120-V lightbulb and a 60-W, 120-V lightbulb

are connected in two different ways as shown. In each

case, which bulb glows more brightly? Ignore change

of filament resistance with current (and temperature).



Parallel Example: Current in one branch.

What is the current through the 500-Ω resistor

shown?



Parallel

Example 26-8:

Analyzing a circuit.

A 9.0-V battery whose

internal resistance r is

0.50 Ω is connected in

the circuit shown. (a)

How much current is

drawn from the

battery? (b) What is

the terminal voltage of

the battery?

Note: slight error in figure and text

chapter 25 electric currents and resistancepeople.virginia.edu/~ben/2415131/lecture_8.pdf ·...

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