2/25 Do now Sphere A has a positive charge of 9 C. Sphere B and sphere C are neutral. Sphere A is...
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Transcript of 2/25 Do now Sphere A has a positive charge of 9 C. Sphere B and sphere C are neutral. Sphere A is...
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 ResistanceKnow: - 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.
4.2.2 Electrical Resistance
Electrical Stopping Power
Definitions• resistance: OPPOSITION TO CURRENT
• unit: Ω• factors that change resistance:
• resistivity: MATERIAL • length• cross-sectional area• temperature
A
LR
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)
R
A
Resistance Factors
R
ρ
R
L
R
Temp.
To build an “ideal” conductor with the smallest possible resistance you would
select one that is:
LOW RESISTIVITY, SHORT, WIDE, COLD
• Determine the resistance of a 1.0 meter long copper wire with a cross-sectional area of 0.01 meter2.
Example #1
R = ρL / A
R = (1.72 x 10-8 Ω·m)(1.0 m) / (0.01 m2)
R = 1.72 x 10-6 Ω
• 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?
Example #2
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• 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 ______________.
1. halved 2. doubled 3. unchanged 4. 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 – ResistancePURPOSE:
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) R (Ω) Area (m2)Length (m) Resistance Area (Ω m∙ ∙ 2)
ObjectivesKnow:
– 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.
4.2.3A Ohm’s Law & Circuit Basics
What Make Electrons Flow Anyway?
Ohm’s Law• Voltage results in current flow
• More voltage = more current
• Resistance opposes current flow• More resistance = less current
R
VI
Resistance: R = V / I
V
I
Slope is resistance
• R is the slope of a potential difference vs. current graph. The resistance is a constant for a metallic conductor at constant temperature.
V
I
Non-Ohmic magterialOhmic material
• A potential difference of 25.0 volts is supplied to a circuit with 100 ohms of resistance.– How much current flows through this circuit?
Example #1
I = V / R
I = 25.0 V / 100 Ω
I = 0.25 A
• A current of 2.0 amperes flows through a 10 ohm resistance.– What voltage must be applied to this resistance?
Example #2
I = V / R
V = IR
V = (2.0 A)(10 Ω)
V = 20 V
• A 10 volt battery establishes a current of 5.0 amperes in a circuit.– What is the resistance of this circuit?
Example #3
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
batterycell
Circuit Elements – Resistances
variable resistorfixed resistor
lamp
Circuit Elements – Switch
switch
Circuit Elements – Measuring Devices
Measures: VOLTAGEResistance: HIGH
Connect to circuit: OUTSIDE
voltmeter
Measures CURRENTResistance: LOW
Connect to circuit: INSIDE
ammeter
Measurements
V = 5V
R = 2.5Ω
V
V V
V
5V
0V
5V
0VA2A
A2A
Voltmeter measuresRELATIVE Potential
differencesfrom OUTSIDE
the circuit
Ammeter measures
Curretn flow INSIDE
the circuit
Graphs: I vs. V and I vs. R
I
V
I
R
Current and resistance have an inverse relationship
Current and potential difference have a direct relationship. The slope is equivalent to the reciprocal of the resistance of the resistor.
I vs. V I vs. R
Rslope
1
R
VI
Ohm's Law as a Predictor of Current
• 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.
R
VI
Check Your Understanding1. 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 resistancec. increase the voltaged. increase the resistance
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 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
a. current, only b. resistance, only c. both current and resistance d. 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
Check Your Understanding
Lab 15 – Resistance
PURPOSE: 1. Determine the relationship between Resistance and the length of the wire2. Determine the relationship between Resistance and the area of the wire3. Determine resistivity of the wireMATERIAL: • Nichrome wire boards, multipurpose meter, ruler, graph paper
DATA:
R (Ω) L (m) R (Ω) Area (m2)Length (m) Resistance Area (Ω m∙ ∙ 2)
End of 4.2.3A