Transcript Unit 3 C Current and Resistance Power

Chapter 17
Current and Resistance
Bright Storm on
Electric Current
Read and take notes on
pages 531-534 in
Conceptual Physics Text
Or lesson 2 C Electric Current on Physics Classroom
Read and take notes on
pages 570-571 in
College Physics Text
Current
• Practical applications were based on static
electricity.
• A steady source of electric current allowed
scientists to learn how to control the flow of
electric charges in circuits.
Introduction
Electric Current
• The current is the rate at which the charge flows
through a surface.
– Look at the charges flowing perpendicularly through a
surface of area A.
• The SI unit of current is Ampere (A)
– 1 A = 1 C/s
Section 17.1
Instantaneous Current
• The instantaneous current is the limit of the average
current as the time interval goes to zero:
• If there is a steady current, the average and
instantaneous currents will be the same.
• SI unit: A
Section 17.1
Electric Current, Cont.
• The direction of the current is the
direction positive charge would
flow.
– This is known as conventional
current direction.
• In a common conductor, such as
copper, the current is due to the
motion of the negatively
charged electrons.
• It is common to refer to a moving
charge as a mobile charge carrier.
– A charge carrier can be positive
or negative.
Section 17.1
EXAMPLE 17.1 Turn On the Light
Goal Apply the concept of current.
Problem The amount of charge that passes through the filament of a
certain lightbulb in 2.00 s is 1.67 C. Find (a) the average current in the
lightbulb and (b) the number of electrons that pass through the filament in
5.00 s.
Strategy Substitute into Equation 17.1a for part (a), then multiply the
answer by the time given in part (b) to get the total charge that passes in
that time. The total charge equals the number N of electrons going through
the circuit times the charge per electron.
SOLUTION
(a) Compute the average current in the lightbulb.
Substitute the charge and time to find the current:
ΔQ 1.67 C
=
= 0.835 A
Iav =
Δt 2.00 s
(b) Find the number of electrons passing through the filament in 5.00 s.
The total number N of electrons times the charge per electron equals the
total charge, IavΔt.
(1) Nq = IavΔt
Substitute and solve for N.
N(1.60 10-19 C/electron) = (0.835 A)(5.00 s)
N = 2.61 1019 electrons
LEARN MORE
Remarks In developing the solution, it was important to use units to ensure
the correctness of equations such as Equation (1). Notice the enormous
number of electrons passing through a given point in a typical circuit.
Question In comparing wires of identical shape and carrying the same
current but made of different metals, is the drift velocity in each wire
inversely proportional to the number density of charge carriers?
Yes. To have the same current from k times as many carriers, it takes k
times the drift speed.
Yes. To have the same current from k times as
many carriers, it takes 1/k times the drift speed.
No. The drift velocity
is directly proportional to the number density of carriers.
velocity is the same for all conductors.
No. The drift
Read and take notes on
pages 572-574 in
College Physics Text
Current and Drift Speed
• Charged particles move
through a conductor of
cross-sectional area A.
• n is the number of charge
carriers per unit volume.
• n A Δx is the total number
of charge carriers.
Section 17.2
Current and Drift Speed, Cont.
• The total charge is the number of carriers times the
charge per carrier, q
– ΔQ = (n A Δx) q
• The drift speed, vd, is the speed at which the carriers
move.
– vd = Δx/ Δt
• Rewritten: ΔQ = (n A vd Δt) q
• Finally, current, I = ΔQ/Δt = nqvdA
Section 17.2
Current and Drift Speed, Final
• If the conductor is isolated, the electrons
undergo random motion.
• When an electric field is set up in the
conductor, it creates an electric force on the
electrons and hence a current.
Section 17.2
Charge Carrier Motion in a Conductor
• The zig-zag black line
represents the motion of a
charge carrier in a
conductor.
– The net drift speed is small.
• The sharp changes in
direction are due to
collisions.
• The net motion of electrons
is opposite the direction of
the electric field.
Section 17.2
Electrons in a Circuit
• Assume you close a switch to turn on a light.
• The electrons do not travel from the switch to the
bulb.
• The electrons already in the bulb move in response
to the electric field set up in the completed circuit.
• A battery in a circuit supplies energy (not charges) to
the circuit.
Electrons in a Circuit, Cont.
• The drift speed is much smaller than the average
speed between collisions.
• When a circuit is completed, the electric field travels
with a speed close to the speed of light.
• Although the drift speed is on the order of 10-4 m/s,
the effect of the electric field is felt on the order of
108 m/s.
Section 17.2
Circuits
• A circuit is a closed path of some sort around
which current circulates.
• A circuit diagram can be used to represent the
circuit.
• Quantities of interest are generally current
and potential difference.
Section 17.3
Read and take notes on
pages 574-575 in
College Physics Text
Meters in a Circuit – Ammeter
• An ammeter is used to measure current.
– In line with the bulb, all the charge passing through the
bulb also must pass through the meter.
Section 17.3
Meters in a Circuit – Voltmeter
• A voltmeter is used to measure voltage (potential
difference).
– Connects to the two contacts of the bulb
Section 17.3
Link to Homework Questions on
Web Assign
Unit 3 C Current and Resistance
# 1-3
Georg Simon Ohm
• 1787 – 1854
• Formulated the concept
of resistance
• Discovered the
proportionality
between current and
voltages
Section 17.4
Bright Storm on
Ohm’s Law
Read and take notes on
pages 534-538 in
Conceptual Physics Text
Or lesson 3 b and 3 c on Physics Classroom
Read and take notes on
pages 575-576 in
College Physics Text
Resistance
• In a conductor, the voltage applied across the
ends of the conductor is proportional to the
current through the conductor.
• The constant of proportionality is the
resistance of the conductor.
Section 17.4
Resistance, Cont.
• Units of resistance are ohms (Ω)
–1Ω=1V/A
• Resistance in a circuit arises due to collisions
between the electrons carrying the current
with the fixed atoms inside the conductor.
Section 17.4
Ohm’s Law
• Experiments show that for many materials, including
most metals, the resistance remains constant over a
wide range of applied voltages or currents.
• This statement has become known as Ohm’s Law.
– ΔV = I R
• Ohm’s Law is an empirical relationship that is valid
only for certain materials.
– Materials that obey Ohm’s Law are said to be ohmic.
Section 17.4
Ohm’s Law, Cont.
• An ohmic device
• The resistance is constant
over a wide range of
voltages.
• The relationship between
current and voltage is linear.
• The slope is related to the
resistance.
Section 17.4
Ohm’s Law, Final
• Non-ohmic materials are
those whose resistance
changes with voltage or
current.
• The current-voltage
relationship is nonlinear.
• A diode is a common
example of a non-ohmic
device.
Section 17.4
Read and take notes on
pages 577-578 in
College Physics Text
Resistivity
• The resistance of an ohmic conductor is proportional
to its length, L, and inversely proportional to its
cross-sectional area, A.
– ρ is the constant of proportionality and is called the
resistivity of the material.
– See table 17.1
Section 17.4
EXAMPLE 17.3 The Resistance of Nichrome Wire
Goal Combine the concept of resistivity with Ohm's law.
Problem (a) Calculate the resistance per unit length of a 22-gauge
Nichrome wire of radius 0.321 mm. (b) If a potential difference of 10.0 V
is maintained across a 1.00-m length of the Nichrome wire, what is the
current in the wire? (c) The wire is melted down and recast with twice its
original length. Find the new resistance RN as a multiple of the old
resistance RO.
Strategy Part (a) requires substitution into Equation 17.5 after calculating
the cross-sectional area, whereas part (b) is a matter of substitution into
Ohm's law. Part (c) requires some algebra. The idea is to take the
expression for the new resistance and substitute expressions for N and AN,
the new length and cross-sectional area, in terms of the old length and
cross-section. For the area substitution, remember that the volumes of the
old and new wires are the same.
SOLUTION
(a) Calculate the resistance per unit length.
Find the cross-sectional area of the wire:
A = πr2 = π(3.210 10-4 m)2 = 3.24 10-7 m2
Obtain the resistivity of Nichrome, solve for R/ , and substitute:
R ρ 1.5 10-6 Ω · m
= =
= 4.6 Ω/m
-7
2
A 3.24 10 m
(b) Find the current in a 1.00-m segment of the wire if the potential
difference across it is 10.0 V:
Substitute given values into Ohm's law:
I=
ΔV
R
=
10.0 V
4.6 Ω
= 2.2 A
(c) If the wire is melted down and recast with twice its original length, find
the new resistance as a multiple of the old.
Find the new area AN in terms of the old area AO, using the fact the volume
doesn't change and N = 2 O.
VN = VO → AN N = AO
O
→ AN = AO (
AN = AO( O/2 O) = AO/2
Substitute to find the new resistance:
RN =
ρ N ρ(2 O)
ρO
=
=4
= 4RO
AN (AO/2)
AO
O
/ N)
LEARN MORE
Remarks The resistivity of Nichrome is about 100 times that of copper, a
typical good conductor. Therefore, a copper wire of the same radius would
have a resistance per unit length of only 0.052 Ω/m, and a 1.00-m length of
copper wire of the same radius would carry the same current (2.2 A) with
an applied voltage of only 0.115 V.
Because of its resistance to oxidation, Nichrome is often used for heating
elements in toasters, irons, and electric heaters.
Question Would replacing the Nichrome with copper wire of the same size
result in a higher current or lower current?
the same current
lower current
higher current
Read and take notes on
pages 579 in
College Physics Text
Temperature Variation of Resistivity
• For most metals, resistivity increases with
increasing temperature.
– With a higher temperature, the metal’s
constituent atoms vibrate with increasing
amplitude.
– The electrons find it more difficult to pass through
the atoms.
Section 17.5
Temperature Variation of Resistivity, Cont.
• For most metals, resistivity increases approximately
linearly with temperature over a limited temperature
range.
– ρ is the resistivity at some temperature T
– ρo is the resistivity at some reference temperature To
• To is usually taken to be 20° C
–  is the temperature coefficient of resistivity
Section 17.5
Temperature Variation of Resistance
• Since the resistance of a conductor with
uniform cross sectional area is proportional to
the resistivity, you can find the effect of
temperature on resistance.
Section 17.5
EXAMPLE 17.4 A Platinum Resistance Thermometer
Goal Apply the temperature dependence of resistance.
Problem A resistance thermometer, which measures temperature by measuring the change in
resistance of a conductor, is made of platinum and has a resistance of 50.0 Ω at 20.0°C. (a)
When the device is immersed in a vessel containing melting indium, its resistance increases to
76.8 Ω. From this information, find the melting point of indium. (b) The indium is heated further
until it reaches a temperature of 235°C. What is the ratio of the new current in the platinum to the
current Imp at the melting point?
Strategy For part (a), solve for T - T0 and get α for platinum, substituting known quantities. For
part (b), use Ohm's law.
SOLUTION
(a) Find the melting point of indium:
Solve for T - T0.
76.8 Ω - 50.0 Ω
R - R0
=
= 137°C
T - T0 =
-3
-1
αR0 [3.92 10 (°C) ][50.0 Ω]
Substitute T0 = 20.0°C and obtain the melting point of indium:
T = 157°C
(b) Find the ratio of the new current to the old when the temperature rises
from 157°C to 235°C.
Write the equation with R0 and T0 replaced by Rmp and Tmp, the resistance
and temperature at the melting point.
R = Rmp[1 + α(T-Tmp)]
According to Ohm's law, R = ΔV/I and Rmp = ΔV/Imp. Substitute these
expressions into the equation.
ΔV
I
=
ΔV
Imp
[1 +α(T - Tmp)]
Cancel the voltage differences, invert the two expressions, and then divide
both sides by Imp:
I
Imp
=
1
1 +α(T - Tmp)
Substitute T = 235°C, Tmp = 157°C, and the value for α, obtaining the
desired ratio:
I
= 0.766
Imp
LEARN MORE
Remarks As the temperature rises, both the rms speed of the electrons in
the metal and the resistance increase.
Question What happens to the drift speed of the electrons as the
temperature rises? (Select all that apply.)
It becomes smaller, so that the resistance increases and current
decreases.
It becomes larger, so that the resistance decreases and current
increases.
It remains the same.
It becomes smaller because the
atoms of the metal vibrate more strongly and interfere more with electron
drift motion.
It becomes larger because the electrons become more
energetic and move faster.
Link to Homework Questions on
Web Assign
Unit 3 C Current and Resistance
# 4-7
Read and take notes on
pages 580-582 in
College Physics Text
Electrical Energy in a Circuit
• In a circuit, as a charge moves through the battery,
the electrical potential energy of the system is
increased by ΔQΔV.
– The chemical potential energy of the battery decreases by
the same amount.
• As the charge moves through a resistor, it loses this
potential energy during collisions with atoms in the
resistor.
– The temperature of the resistor will increase.
Section 17.6
Energy Transfer in the Circuit
• Consider the circuit
shown.
• Imagine a quantity of
positive charge, DQ,
moving around the
circuit from point A
back to point A.
Section 17.6
Energy Transfer in the Circuit, Cont.
• Point A is the reference point.
– It is grounded and its potential is taken to be zero.
• As the charge moves through the battery from
A to B, the potential energy of the system
increases by DQDV.
– The chemical energy of the battery decreases by
the same amount.
Section 17.6
Energy Transfer in the Circuit, Final
• As the charge moves through the resistor, from C to
D, it loses energy in collisions with the atoms of the
resistor.
• The energy is transferred to internal energy.
• When the charge returns to A, the net result is that
some chemical energy of the battery has been
delivered to the resistor and caused its temperature
to rise.
Section 17.6
Bright Storm on
Power
Power
• In a conductor carrying a current, the electric
potential of the charges is continually
decreasing.
• Positive charges move from regions of high
potential to regions of low potential.
• ΔUcharges = q ΔV is negative
– Often only the magnitude is desired
• The power delivered to the circuit element is
the energy divided by the elapsed time.
Section 17.1
Electrical Energy and Power, Cont.
• The rate at which the energy is lost is the
power.
• From Ohm’s Law, alternate forms of power
are
Section 17.6
Electrical Energy and Power, Final
• The SI unit of power is Watt (W).
– I must be in Amperes, R in ohms and DV in Volts
• The unit of energy used by electric companies
is the kilowatt-hour.
– This is defined in terms of the unit of power and
the amount of time it is supplied.
– 1 kWh = 3.60 x 106 J
Section 17.6
EXAMPLE 17.5 The Cost of Lighting Up Your Life
Goal Apply the electric power concept and calculate the cost of power
usage using kilowatt-hours.
Problem A circuit provides a maximum current of 20.0 A at an operating
voltage of 1.20 102 V. (a) How many 75 W bulbs can operate with this
voltage source? (b) At $0.120 per kilowatt-hour, how much does it cost to
operate these bulbs for 8.00 h?
Strategy Find the necessary power with = IΔV then divide by 75.0 W per
bulb to get the total number of bulbs. To find the cost, convert power to
kilowatts and multiply by the number of hours, then multiply by the cost
per kilowatt-hour.
SOLUTION
(a) Find the number of bulbs that can be lighted.
Substitute to get the total power.
102 V) = 2.40 103 W
total = IΔV = (20.0 A)(1.20
Divide the total power by the power per bulb to get the number of bulbs.
2.40 103 W
total
Number of bulbs =
=
= 32.0
75.0 W
bulb
(b) Calculate the cost of this electricity for an 8.00-h day.
Find the energy in kilowatt-hours.
1.00 kW
Energy = t = (2.40 10 W) 1.00 103 W (8.00 h)
3
(
)
= 19.2 kWh
Multiply the energy by the cost per kilowatt-hour.
Cost = (19.2 kWh)($0.12/kwH) = $2.30
LEARN MORE
Remarks This amount of energy might correspond to what a small office
uses in a working day, taking into account all power requirements (not just
lighting). In general, resistive devices can have variable power output,
depending on how the circuit is wired. Here, power outputs were specified,
so such considerations were unnecessary.
Question Considering how hot the parts of an incandescent light bulb get
during operation, guess what fraction of the energy emitted by an
incandescent light bulb is in the form of visible light.
80%
50%
10%
EXAMPLE 17.6 The Power Converted by an Electric Heater
Goal Calculate an electrical power output and link to its effect on the
environment through the first law of thermodynamics.
Problem An electric heater is operated by applying a potential difference
of 50.0 V to a Nichrome wire of total resistance 8.00 Ω. (a) Find the
current carried by the wire and the power rating of the heater. (b) Using
this heater, how long would it take to heat 2.50 103 moles of diatomic gas
(e.g., a mixture of oxygen and nitrogen, or air) from a chilly 10.0°C to
25.0°C? Take the molar specific heat at constant volume of air to be (5/2)R.
Strategy For part (a), find the current with Ohms law and substitute into
the expression for power. Part (b) is an isovolumetric process, so the
thermal energy provided by the heater all goes into the change in internal
energy, ΔU. Calculate this quantity using the first law of thermodynamics
and divide by the power to get the time.
SOLUTION
(a) Compute the current and power output.
Apply Ohms law to get the current.
ΔV 50.0 V
=
= 6.25 A
I=
R 8.00 Ω
Substitute to find the power.
= I2R = (6.25 A)2(8.00 Ω) = 313 W
(b) How long does it take to heat the gas?
Calculate the thermal energy transfer from the first law. Note that W = 0
because the volume doesn't change
Q = ΔU = nCvΔT = (2.50 103 mol)((5/2) · 8.31 J/mol · K)(298 K − 283
K)
= 7.79 105 J
Divide the thermal energy by the power to get the time.
t=
Q 7.79 105 J
=
313 W
= 2.49 105 s
LEARN MORE
Remarks The number of moles of gas given here is approximately what
would be found in a bedroom. Warming the air with this space heater
requires only about 40 minutes. The calculation, however, doesn't take into
account conduction losses. Recall that a 20-cm-thick concrete wall
permitted the loss of more than 2 megajoules an hour by conduction!
Question If the heater wire is replaced by a wire with lower resistance, the
time required to heat the gas is:
increased.
unchanged.
decreased.
Read and take notes on
pages 584-587 in
College Physics Text
Superconductors
• A class of materials and
compounds whose
resistances fall to virtually
zero below a certain
temperature, TC
– TC is called the critical
temperature
• The graph is the same as a
normal metal above TC, but
suddenly drops to zero at TC
Section 17.7
Superconductors, Cont.
• The value of TC is sensitive to
– Chemical composition
– Pressure
– Crystalline structure
• Once a current is set up in a superconductor, it
persists without any applied voltage.
– Since R = 0
Section 17.7
Superconductor Timeline
• 1911
– Superconductivity discovered by H. Kamerlingh Onnes
• 1986
– High temperature superconductivity discovered by Bednorz and
Müller
– Superconductivity near 30 K
• 1987
– Superconductivity at 96 K and 105 K
• Current
– Superconductivity at 150 K
– More materials and more applications
Section 17.7
Superconductor, Final
• Good conductors do not
necessarily exhibit
superconductivity.
• One application is the
construction of
superconducting
magnets.
Section 17.7
Electrical Activity in the Heart
• Every action involving the body’s muscles is
initiated by electrical activity.
• Voltage pulses cause the heart to beat.
• These voltage pulses are large enough to be
detected by equipment attached to the skin.
Section 17.8
Operation of the Heart
• The sinoatrial (SA) node
initiates the heartbeat.
• The electrical impulses
cause the right and left
artial muscles to contract.
• When the impulse reaches
the atrioventricular (AV)
node, the muscles of the
atria begin to relax.
• The ventricles relax and the
cycle repeats.
Section 17.8
Electrocardiogram (EKG)
• A normal EKG
• P occurs just before the
atria begin to contract.
• The QRS pulse occurs in the
ventricles just before they
contract.
• The T pulse occurs when
the cells in the ventricles
begin to recover.
Section 17.8
Link to Homework Questions on
Web Assign
Unit 3 C Current and Resistance
# 8-10
Link to Discussion Questions on
Web Assign
Unit 3 C Current and Resistance
Grading Rubric for Unit 3 C Current and Resistance
Name: ______________________
Conceptual Physics Text Notes
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