Lesson 2: Electrical Resistance

Download Report

Transcript Lesson 2: Electrical Resistance

2/25 Do now
• Sphere A has a positive charge of 9 C. Sphere B and sphere C are
neutral. Sphere A is used to charge sphere B by conduction.
Sphere A is used to charge sphere C by induction.
1. How many elementary charges does sphere A have before it is
used to charge spheres B and C?
2. What is the type of charge of sphere B after charging?
3. What is the type of charge of sphere C after charging?
Due:
19.2 notes
Assignment:
Castle learning
Unit test:
Questions from packets
Lesson 2: Electrical Resistance
Know:
- Definition and equation for electrical resistance.
Understand
- The electrical resistance of an object depends on the resistivity of the
material it is made from; its physical measurements (cross-sectional area
and length); and its temperature.
Be able to
- Calculate electrical resistance; resistivity; length; or cross-sectional area of
an object.
- Select graphs show the relationship between electrical resistance and:
length; cross-sectional area; resistivity; and/or temperature.
- Use resistivity to determine the composition of an object.
- Explain the difference between resistance and resistivity.
- Determine the effect of changing the length; cross-sectional area; and/or
temperature on the resistance of an object.
Electrical Stopping
Power
4.2.2 Electrical Resistance
Definitions
• resistance: OPPOSITION TO CURRENT
• unit: Ω
• factors that change resistance:
• resistivity: MATERIAL
• length
• cross-sectional area
• temperature
R 
L
A
L - the length of the wire (meters),
A - the cross-sectional area of the wire (m2),
ρ - the resistivity of the material (in Ω•meter).
R - the resistance of the wire (in Ω)
Resistance Factors
R
R
To build an “ideal” conductor with the
smallestρpossible resistance you
A would
select one that is:
LOW RESISTIVITY, SHORT, WIDE, COLD
R
R
L
Temp.
Example #1
• Determine the resistance of a 1.0 meter long copper
wire with a cross-sectional area of 0.01 meter2.
R = ρL / A
R = (1.72 x 10-8 Ω·m)(1.0 m) / (0.01 m2)
R = 1.72 x 10-6 Ω
Example #2
• A piece of wire that has a length of 5.0 x 107 meters
and a cross-sectional area of 0.025 meter2 has a
resistance of 31.8 ohms.
– What is the composition of this wire?
R = ρL / A
ρ = RA / L
ρ = (31.8 Ω)(0.025 m2) / (5.0 x 107 m)
ρ = 1.59 x 10-8 Ω·m
example
•
An incandescent light bulb is supplied with a
constant potential difference of 120 volts. As the
filament of the bulb heats up,
1. What happens to the resistance?
2. What happens to the current?
example
•
1.
2.
3.
4.
If the cross-sectional area of a metallic
conductor is halved and the length of the
conductor is doubled, the resistance of the
conductor will be ______________.
halved
doubled
unchanged
quadrupled
example
•
A 12.0-meter length of copper wire has a resistance
of 1.50 ohms. How long must an aluminum wire
with the same cross-sectional area be to have the
same resistance?
example
• Pieces of aluminum, copper, gold, and silver
wire each have the same length and the same
cross-sectional area. Which wire has the
lowest resistance at 20°C?
End of 4.2.2 - PRACTICE
2/26 Do now
• An object has +2 C of charge is placed in an electric field E,
experiencing electric for Fe, has an electric potential V and
electric potential energy W.
1. What is the order of magnitude for the charge on this object in
elementary charges?
• If the charge is doubled,
2. what is going to happen to E?
3. what is going to happen to V?
4. what is going to happen to Fe?
5. what is going to happen to W?
Ω Work
1. Castle learning
2. Note 19.3 – due tomorrow
3. Project – poster or power point on one topic in this
chapter - due 3/3
4. Chapter test on Tues. 3/4 – include some static electricity
questions from last chapter. Extra credit: make your own
Ω joke – due Mon 3/3
5. No Post Session Today
6. Unit test – 3/18: Questions from packets
Lab 15 – Resistance
PURPOSE:
1. Determine the relationship between Resistance and the length of the wire
2. Determine the relationship between Resistance and the area of the wire
3. Determine resistivity of the wire
MATERIAL:
• Nichrome wire boards, multipurpose meter, ruler, graph paper
DATA:
R (Ω)
L (m)
Length (m)
Resistance ∙Area
(Ω∙m2)
R (Ω)
Area (m2)
Objectives
Know:
– Equation for Ohm’s Law.
Understand
– Current is directly proportional to voltage and inversely
proportional to electrical resistance.
Be able to
– Determine current; voltage; resistance; and/or power in a
system with a single resistor.
- Sketch/interpret graphs of relating voltage; current;
resistance;
- Determine whether or not a particular object obeys Ohm’s
Law.
What Make Electrons
Flow Anyway?
4.2.3A Ohm’s Law & Circuit Basics
Ohm’s Law
• Voltage results in current flow
• More voltage = more current
• Resistance opposes current flow
• More resistance = less current
I 
V
R
Resistance: R = V / I
• R is the slope of a potential difference vs. current
graph. The resistance is a constant for a metallic
conductor at constant temperature.
V
V
Slope is resistance
I
Ohmic material
I
Non-Ohmic magterial
Example #1
• A potential difference of 25.0 volts is supplied to a
circuit with 100 ohms of resistance.
– How much current flows through this circuit?
I=V/R
I = 25.0 V / 100 Ω
I = 0.25 A
Example #2
• A current of 2.0 amperes flows through a 10 ohm
resistance.
– What voltage must be applied to this resistance?
I=V/R
V = IR
V = (2.0 A)(10 Ω)
V = 20 V
Example #3
• A 10 volt battery establishes a current of 5.0 amperes
in a circuit.
– What is the resistance of this circuit?
I=V/R
R=V/I
R = (10 A) / (5.0 A)
R = 2.0 Ω
What is a circuit?
• A continuous loop through which current flows
from an area of high voltage to a an area of low
voltage.
Circuit Elements – Voltage Sources
cell
battery
Circuit Elements – Resistances
fixed resistor
variable resistor
lamp
Circuit Elements – Switch
switch
Circuit Elements – Measuring Devices
voltmeter
ammeter
Measures: VOLTAGE
Resistance: HIGH
Connect to circuit: OUTSIDE
Measures CURRENT
Resistance: LOW
Connect to circuit: INSIDE
Measurements
V
5V
R = 2.5Ω
V
2A
0V
Voltmeter
measures
RELATIVE
Potential
differences
from OUTSIDE
the circuit
2A
A
A
V = 5V
V
5V
0V
V
Ammeter
measures
Curretn flow
INSIDE
the circuit
Graphs: I vs. V and I vs. R
I 
I vs. V
V
I vs. R
R
I
slope 
1
I
R
V
Current and potential
difference have a direct
relationship. The slope is
equivalent to the reciprocal of
the resistance of the resistor.
R
Current and
resistance have an
inverse relationship
Ohm's Law as a Predictor of Current
I 
V
R
• The current in a circuit is directly proportional to the
electric potential difference impressed across its ends
and inversely proportional to the total resistance offered
by the external circuit.
• The greater the battery voltage (i.e., electric potential
difference), the greater the current. a twofold increase in
the battery voltage would lead to a twofold increase in
the current (if all other factors are kept equal).
• The greater the resistance, the less the current. An
increase in the resistance of the load by a factor of two
would cause the current to decrease by a factor of two to
one-half its original value.
Check Your Understanding
1. Which of the following will cause the current
through an electrical circuit to decrease?
Choose all that apply.
a. decrease the voltage
b. decrease the resistance
c. increase the voltage
d. increase the resistance
Check Your Understanding
2. A copper wire is connected across a constant
voltage source. The current flowing in the
wire can be increased by increasing the
wire's
a. cross-sectional area
b. length
c. resistance
d. temperature
Check Your Understanding
3. A series circuit has a total resistance of 1.00 × 102
ohms and an applied potential difference of 2.00 ×
102 volts. What is the amount of charge passing any
point in the circuit in 2.00 seconds?
Check Your Understanding
4. A long copper wire was connected to a voltage
source. The voltage was varied and the current
through the wire measured, while temperature was
held constant. Using the graph to find the resistance
of the copper wire.
Check Your Understanding
• A student conducted an experiment to determine the resistance
of a light bulb. As she applied various potential differences to the
bulb, she recorded the voltages and corresponding currents and
constructed the graph below. The student concluded that the
resistance of the light bulb was not constant.
5.
What evidence from the graph supports the student’s
conclusion?
6.
According to the graph, as the
potential difference increased,
what happens to the
resistance of the light bulb?
Check Your Understanding
7.
a.
b.
c.
d.
A circuit consists of a resistor and a battery. Increasing the
voltage of the battery while keeping the temperature of the
circuit constant would result in an increase in
current, only
resistance, only
both current and resistance
neither current nor resistance
Check Your Understanding
8. Sketch a graph that best represents the relationship
between the potential difference across a metallic
conductor and the electric current through the
conductor
a. At constant temperature T1
b. At a higher constant temperature T2.
V
I
Check Your Understanding
9. A 1.5-volt, AAA cell supplies 750 milliamperes of
current through a flashlight bulb for 5.0 minutes,
while a 1.5-volt, C cell supplies 750 milliamperes of
current through the same flashlight bulb for 20.
minutes. Compared to the total charge transferred
by the AAA cell through the bulb, the total charge
transferred by the C cell through the bulb is
a. half as great
b. twice as great
c. the same
d. four times as great
Lab 15 – Resistance
PURPOSE:
1. Determine the relationship between Resistance and the length of the wire
2. Determine the relationship between Resistance and the area of the wire
3. Determine resistivity of the wire
MATERIAL:
• Nichrome wire boards, multipurpose meter, ruler, graph paper
DATA:
R (Ω)
L (m)
Length (m)
Resistance ∙Area
(Ω∙m2)
R (Ω)
Area (m2)
End of 4.2.3A