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2/27 Do now Sphere A has a charge of -9.0 x 10 24 e. 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. What is the order of magnitude of charges in Coulombs on sphere A? 2. What is the type of charge on sphere B after charging? 3. What is the type of charge on sphere C after charging? Due: 19.3 notes

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2/27 Do now. Sphere A has a charge of -9.0 x 10 24 e. 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 . What is the order of magnitude of charges in Coulombs on sphere A? - PowerPoint PPT Presentation

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Page 1: 2/27 Do now

2/27 Do now• Sphere A has a charge of -9.0 x 1024 e. 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. What is the order of magnitude of charges in Coulombs on sphere A?

2. What is the type of charge on sphere B after charging? 3. What is the type of charge on sphere C after charging?

Due:19.3 notes

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Ω Work1. Castle learning

2. Project – poster or power point on one topic in this chapter - due 3/3

3. 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

4. Unit test – 3/14: Questions from packets

5. Chapter 20-22 reading assignment

6. Project – choose one – due Fri 3/14

– build a simple motor

– build a series or a parallel circuit with mini lights bulbs, insulated wires, switches and batteries.

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Lesson 4: Electrical PowerKnow: • Definition and equation for electrical power.

Understand • Power is directly proportional to both voltage and current.

Be able to • Determine power in a system with a single resistor. • Sketch/interpret graphs of relating voltage; current; resistance

and power with each other (assuming that all other variables are fixed.)

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Power: Putting Charges to Work Electrical devices, generally referred to as loads, have power ratings.

A 1200 W hair dryer indicates it transfers 1200 Joules of electrical energy to heat, wind, sound energy in 1 second.

• A circuit with a battery and a wire leading from positive to negative terminal without a load would lead to a high rate of charge flow. Such a circuit is referred to as a short circuit. It would heat the wires to a high temperature and drain the battery of its energy rather quickly.

timeEnergyPower

The unit of power is watt. 1 watt = 1 joule / second

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4.2.3B Electrical Power

What About Watts?

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Power Law• Moving electrons (current) requires ENERGY

• How much energy gets used depends on:• Strength of push – VOLTAGE• Rate of flow – CURRENT

IVP RVI RIRVIVP 2

2

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• A 12 volt battery is connected to a circuit which allows 10 amperes of current to flow.– What is the power output of this circuit?

Example #1

P = IV

P = (12 V)(10 A)

P = 120 W

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• A 100 watt light bulb is connected to a 120 volt power supply.– What amount of current must pass through the light bulb?

Example #2

P = IV

100 W = (120 V) I

I = 0.833 A

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• A 2.0 ampere current passes through a circuit with a 300 ohm resistance.– What is the power generated in this circuit?

Example #3

P = I2 R

P = (2.0 A)2 (300 Ω)

P = 1200 W or 1.2 kW

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P = I2•R relate current and resistance to power, notice double importance of current. Unit: A2∙Ω

P = V2/R relate potential difference and resistance to power, notice double importance of potential difference. Unit: V2/Ω

P = V·I relate potential difference and current to power. Notice that both have equal importance. Unit: V∙A

Warning:

While these three equations provide one with convenient formulas for calculating unknown quantities in physics problems, one must be careful to not misuse them by ignoring conceptual principles regarding circuits.

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Check your understanding1. If a 60-watt bulb in a household lamp was replaced

with a 120-watt bulb, then how many times greater would the current be in that lamp circuit?

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2. Which is a unit of electrical power?a. volt/ampere b. ampere/ohm c. ampere2/ohm d. volt2/ohm

Check your understanding

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Graphs of power vs. R, I, V• P = VI = I2R = V2/R• When V is constant: P = VI; P = V2/R – common house hold

appliances

• When R is constant: P = I2R; P = V2/R – same appliances

P

I

P

RV is slope

Inverse, high R, low P

P P

I VDirect squared Direct squared

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3. As the resistance of a constant-voltage circuit is increased, the power developed in the circuit

a. decreases b. increases c. remains the same

Check your understanding

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4. The potential difference applied to a circuit element remains constant as the resistance of the element is varied. Graph power (P) vs. resistance (R) for this circuit.

P

R

Check your understanding

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5. Graph the relationship between the electrical power and the current in a resistor that obeys Ohm’s Law.

P

I

Check your understanding

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6. An electric motor uses 15 amperes of current in a 440-volt circuit to raise an elevator weighing 11,000 Newtons. What is the average speed attained by the elevator?

Check your understanding

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example

7. To increase the brightness of a desk lamp, a student replaces a 60-watt light bulb with a 100-watt bulb. Compared to the 60-watt bulb, the 100-watt bulb has

a. less resistance and draws more current b. less resistance and draws less current c. more resistance and draws more current d. more resistance and draws less current

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2/28 Do now

• A lithium ion has 3 protons and 4 neutrons. What is the charge in coulombs of the lithium ion?

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Ω Work

1. Castle learning

2. Project – poster or power point on one topic in this chapter - due 3/3

3. 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

4. Unit test – 3/14: Questions from packets

If there is no school Monday, additional castle learning will be assigned. Chapter test will still be on Tuesday!

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8. Which would be thicker (wider) - the filament of a 60-Watt light bulb or the filament of a 100-W light bulb? Explain.

 

 9. Calculate the resistance and the current of a 7.5-Watt night

light bulb plugged into a US household outlet (120 V).  

Check Your Understanding

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Electrical energy

• E = P∙t = V∙I∙t = I2∙R∙t = (V2/R)∙t • The SI unit for energy is Joule.• 1 joule = (1 Newton)(1 meter)

= (1 kg∙m/s2)(1 meter)= 1 kg∙m2/s2

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The kilowatt-hour• Electrical utility companies provide energy for homes

charge those homes for the electrical energy they used. A typical bill will contain a charge for the number of kilowatt-hours of electricity which were consumed.

• How many Joules is in one kWh?

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1. Your 60-watt light bulb is plugged into a 110-volt household outlet and left on for 10 hours. The utility company charges you $0.20 per kWh. What is the cost?

2. A current of 0.40 ampere is measured in a 150 ohm resistor, how much energy is expended by the resistor in 20. seconds?

3. An electric dryer consumes 6.0 × 106 joules of energy when operating at 220 volts for 30. minutes. During operation, how much current does the dryer draws approximately?

Check your understanding

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Energy can be transformed, but is conserved

• The purpose of every circuit is to supply the energy to operate various electrical devices. These devices are constructed to convert the energy of flowing charge into other forms of energy (e.g., light, thermal, sound, mechanical, etc.). Use complete sentences to describe the energy conversions that occur in the following devices.

4. Windshield wipers on a car 5. Defrosting circuit on a car6. Hair dryer

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• Rechargeable batteries has nothing to do with charges. • Rechargeable batteries rely upon a reversible reaction, turning the

chemical products back into chemical reactants within the cell.

Rechargeable Batteries

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Alert: Statement True or False?7. When an electrochemical cell no longer works, it is out of charge

and must be recharged before it can be used again.

8. An electrochemical cell can be a source of charge in a circuit. The charge which flows through the circuit originates in the cell.

9. Charge becomes used up as it flows through a circuit. The amount of charge which exits a light bulb is less than the amount which enters the light bulb.

10. Charge flows through circuits at very high speeds. This explains why the light bulb turns on immediately after the wall switch is flipped.

11. The local electrical utility company supplies millions and millions of electrons to our homes everyday.

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Example 12

• 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?

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Example 13

• Calculate the resistance of a 1.00-kilometer length of nichrome wire with a cross-sectional area of 3.50 × 10-6 meter2 at 20°C.

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3/3 do now• A clothes dryer connected to a 240-volt line

draws 30. amperes of current for 20. minutes (1,200 seconds). Approximately how much electrical energy is consumed by the dryer?[show work]

Due:• Project• Ω jokeΩ work• Packet 4.2.1-4.2.3Test Tomorrow

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Finish lab 15

objectives

Regents Review book pp. 112-113 #29-48• Due today

Presentation of project

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Finish lab 15

3/4 objectives

Regents Review book pp. 112-113 #29-48• Due today

Presentation of project

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3/5 objectives• Due: packet 4.2.1-4.2.3• Test• Homework – 20.1 reading questions due

tomorrow

• Finish Regents Review book pp. 112-113 #29-48• Finish practice in note packet• Finish lab 15

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Quantities, Symbols, Equations and Units!

• The tendency to give attention to units is an essential trait of any good physics student.

• Many of the difficulties associated with solving problems may be traced back to the failure to give attention to units. As more and more electrical quantities and their respective metric units are introduced, it will become increasingly important to organize the information in your head.

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Quantity Symbol EquationsStandard

Metric Unit

Other Units

Potential Difference (a.k.a. voltage)  V    V= W / Q

V = I • R Volt (V) J / C

Current I I = Q / t I = V / R Amperes (A) C / s

V / Ω

Power P

P = W / t P = V∙I

P = V2/RP = I2R

Watt (W)

J / sV∙A

V/ Ω2

A2∙Ω

Resistance R R = ρ•L / A R = V / I Ohm (Ω ) V / A

Energy W W = V • Q W = P • t Joule (J) V • C

W • s

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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)

diameter _________ mArea __________m2

Length _________ m